Police Patrol
First, let's try to find the largest block that two patrols can cover (hint, it's not k). For instance, we can have one patrol cover 1, 2, ..., k. and the other cover 2, 3, ..., k + 1.
So, we can split n houses into blocks of length k + 1. Now, we have to look at what to do with the leftover houses. If there are more than two empty blocks, we need two patrols to cover both blocks. If there is only one empty block, we only need one patrol. Thus, the answer can be computed by the formula in the code below
code
n,k = map(int, raw_input().split())
print (n/(k+1))*2 + min(2, n%(k+1))
Alphabetic Subsequence
We can check that this is equivalent to counting the number of permutations of 0, 1, 2, ..., 9 that appear in s.
There are two solutions:
solution 1
Just do it. This is O(10! * |s|) ~ 3*10^8. This is fast enough in some languages.
code
#include <bits/stdc++.h>
using namespace std;
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define fore(i, b, e) for (int i = (int)(b); i <= (int)(e); ++i)
#define ford(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define pb push_back
#define fi first
#define se second
#define all(x) (x).begin(), (x).end()
typedef vector<int> vi;
typedef pair<int, int> pii;
typedef long long i64;
typedef unsigned long long u64;
typedef long double ld;
typedef long long ll;
const int maxn = 105;
int n;
int a[maxn];
int p[maxn];
bool check() {
int cur = 0;
forn(i, n) {
if (p[a[i]] == cur) ++cur;
}
return cur == 10;
}
int main() {
#ifdef LOCAL
freopen("b.in", "r", stdin);
#endif
char c;
while (cin >> c) { a[n++] = c-'0'; }
forn(i, 10) p[i] = i;
int cnt = 0;
do {
cnt += check();
} while (next_permutation(p, p+10));
cout << cnt << endl;
#ifdef LOCAL
cerr << "Time elapsed: " << clock() / 1000 << " ms" << endl;
#endif
return 0;
}
solution 2
We can do a bitmask dp. We need to be careful not to double count, so we advance our index to the first occurence of some character to take it. See the code for more details.
code
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
const int N = 105;
vector<int> pos[N];
char buf[N];
const int MSK = 1 << 10;
int D[MSK][N];
const int K = 10;
int main() {
scanf("%s", buf);
int n = strlen(buf);
for (int i = 0; i < n; i++) {
pos[buf[i] - '0'].push_back(i);
}
D[0][0] = 1;
for (int msk = 0; msk < MSK - 1; msk++) {
for (int i = 0; i < K; i++) {
if ((msk >> i) & 1) {
continue;
}
for (int j = 0; j <= n + 1; j++) {
if (j == n + 1) {
D[msk | (1 << i)][j] += D[msk][j];
} else {
int p = lower_bound(pos[i].begin(), pos[i].end(), j) - pos[i].begin();
if (p == pos[i].size()) {
p = n + 1;
} else {
p = pos[i][p];
}
D[msk | (1 << i)][p + 1] += D[msk][j];
}
}
}
}
int ans = 0;
for (int i = 0; i <= n; i++) {
ans += D[MSK - 1][i];
}
printf("%d\n", ans);
}
Infinite Graph Game
Consider at each step, instead of adding our score to a global counter, we incremented the score at a particular vertex. So, we only need to simulate one round, and can multiply by an appropriate constant to get the limit value. We can also check the infinite case here (the contribution of a node is nonzero and it never gets chosen so its value never gets reduced).
The problem is that this approach takes O(nk) time in the worst case. Let's look at each edge individually. If node a appears fa times, and node b appears fb times in our sequence, we can get the contribution of this edge onto each node's values in O(fa + fb) time. We can also further reduce this to O(min(fa, fb) * log(max(fa, fb))) time by iterating through the explicit occurences of the node that appears fewer times, and using some data structures to count the number of occurences of the other node between any two explicit occurences.
With this reduction, this brings our running time to O(n * sqrt(n) * log(n)) (for example, see this problem for analysis: http://codeforces.me/problemset/problem/804/D)
c++ code
#include <cstdio>
#include <cassert>
#include <cstdlib>
#include <vector>
using namespace std;
const int MOD = 1000 * 1000 * 1000 + 7;
inline int add(int a, int b) {
if ((a += b) >= MOD) {
a -= MOD;
}
return a;
}
typedef long long llong;
inline int mul(int a, int b) {
return ((llong)a) * b % MOD;
}
inline int sub(int a, int b) {
if ((a -= b) < 0) {
a += MOD;
}
return a;
}
int powmod(int a, int b) {
int res = 1;
while (b > 0) {
if (b & 1) {
res = mul(res, a);
}
a = mul(a, a);
b >>= 1;
}
return res;
}
int inv(int x) {
return powmod(x, MOD - 2);
}
const int N = 100500;
int V[N];
vector<int> history[N];
int AF[N];
void die() {
puts("-1");
exit(0);
}
int pw2[N];
int main() {
int n, m, k;
scanf("%d %d %d", &n, &m, &k);
for (int i = 0; i < n; i++) {
scanf("%d", &V[i]);
}
for (int i = 0; i < k; i++) {
int s;
scanf("%d", &s);
--s;
history[s].push_back(i);
}
pw2[0] = 1;
for (int i = 0; i < k; i++) {
pw2[i + 1] = (pw2[i] % 2 == 0) ? pw2[i] / 2 : (pw2[i] + MOD) / 2;
}
for (int i = 0; i < m; i++) {
int a, b;
scanf("%d %d", &a, &b);
--a, --b;
if (history[a].size() > history[b].size()) {
swap(a, b);
}
if (history[b].empty()) {
continue;
} else if (history[a].empty()) {
if (V[a] == 0) {
continue;
} else {
die();
}
} else {
int prv = 0;
for (int i = 0; i < (int)history[a].size(); i++) {
int x = history[a][i];
int p = lower_bound(history[b].begin(), history[b].end(), x) - history[b].begin();
AF[b] = add(AF[b], pw2[p]);
AF[a] = add(AF[a], mul(p - prv, pw2[i]));
prv = p;
}
AF[a] = add(AF[a], mul((int)history[b].size() - prv, pw2[(int)history[a].size()]));
}
}
int ans = 0;
for (int i = 0; i < n; i++) {
if (AF[i] != 0) {
assert(!history[i].empty());
}
int k = pw2[(int)history[i].size()];
int num = mul(AF[i], V[i]);
int den = sub(1, k);
ans = add(ans, mul(num, inv(den)));
}
printf("%d\n", ans);
}
java code
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.stream.Stream;
import java.io.BufferedWriter;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.util.Collections;
import java.util.ArrayList;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
InfiniteGraphGame solver = new InfiniteGraphGame();
solver.solve(1, in, out);
out.close();
}
static class InfiniteGraphGame {
public int mod = 1000000007;
int[] cc;
int[] sz;
ArrayList<Integer>[] pos;
public void solve(int testNumber, InputReader in, OutputWriter out) {
int n = in.nextInt(), m = in.nextInt(), k = in.nextInt();
int[] v = in.readIntArray(n);
int[] s = in.readIntArray(k);
for (int i = 0; i < k; i++) s[i]--;
pos = Stream.generate(ArrayList::new).limit(n).toArray(ArrayList[]::new);
for (int i = 0; i < n; i++) pos[i].add(0);
cc = new int[k + 1];
sz = new int[n];
for (int i = 1; i <= k; i++) {
cc[i] = pos[s[i - 1]].size();
pos[s[i - 1]].add(i);
sz[s[i - 1]]++;
}
long[] div2 = new long[k + 1];
div2[0] = 1;
long mult = Utils.inv(2, mod);
for (int i = 1; i < div2.length; i++) {
div2[i] = div2[i - 1] * mult % mod;
}
long ans = 0;
for (int i = 0; i < m; i++) {
int a = in.nextInt() - 1, b = in.nextInt() - 1;
if (sz[a] > sz[b]) {
int t = a;
a = b;
b = t;
}
if (sz[b] > 0 && sz[a] == 0 && v[a] > 0) {
out.println(-1);
return;
}
long sa = 0, sb = 0;
int da = 0, db = 0;
int lastw = 0;
for (int q : pos[a]) {
if (q == 0) continue;
int curw = cc[pos[b].get(-Collections.binarySearch(pos[b], q) - 2)];
long w = curw - lastw;
lastw = curw;
sa = (sa + 1L * w * v[a] % mod * div2[da]) % mod;
db += w;
sb = (sb + v[b] * div2[db]) % mod;
da++;
}
{
long w = sz[b] - lastw;
sa = (sa + 1L * w * v[a] % mod * div2[da]) % mod;
db += w;
}
sa = sa * Utils.inv(1 - div2[da] + mod, mod) % mod;
sb = sb * Utils.inv(1 - div2[db] + mod, mod) % mod;
ans = (ans + sa + sb) % mod;
}
out.println(ans);
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void close() {
writer.close();
}
public void println(long i) {
writer.println(i);
}
public void println(int i) {
writer.println(i);
}
}
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1 << 16];
private int curChar;
private int numChars;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int[] readIntArray(int tokens) {
int[] ret = new int[tokens];
for (int i = 0; i < tokens; i++) {
ret[i] = nextInt();
}
return ret;
}
public int read() {
if (this.numChars == -1) {
throw new InputMismatchException();
} else {
if (this.curChar >= this.numChars) {
this.curChar = 0;
try {
this.numChars = this.stream.read(this.buf);
} catch (IOException var2) {
throw new InputMismatchException();
}
if (this.numChars <= 0) {
return -1;
}
}
return this.buf[this.curChar++];
}
}
public int nextInt() {
int c;
for (c = this.read(); isSpaceChar(c); c = this.read()) {
;
}
byte sgn = 1;
if (c == 45) {
sgn = -1;
c = this.read();
}
int res = 0;
while (c >= 48 && c <= 57) {
res *= 10;
res += c - 48;
c = this.read();
if (isSpaceChar(c)) {
return res * sgn;
}
}
throw new InputMismatchException();
}
public static boolean isSpaceChar(int c) {
return c == 32 || c == 10 || c == 13 || c == 9 || c == -1;
}
}
static class Utils {
public static long inv(long N, long M) {
long x = 0, lastx = 1, y = 1, lasty = 0, q, t, a = N, b = M;
while (b != 0) {
q = a / b;
t = a % b;
a = b;
b = t;
t = x;
x = lastx - q * x;
lastx = t;
t = y;
y = lasty - q * y;
lasty = t;
}
return (lastx + M) % M;
}
}
}
Stamp Stamp Stamp
The bounds might look a bit strange, and it was a bit hard to find good bounds for this. Originally, I wanted to set bounds higher so only O(n3·logn) solutions could pass, but it seems the constant factor for that solution is too high. So, I set bounds low enough so a O(n4) solution could pass as well.
O(n^4)
For an O(n4) solution, we can check all O(n^2) substrings. Now, we want to know if this string can stamp out the entire string. We can do this with dp. Let dp[i][j] be true if it is possible to stamp out the prefix 1, ..., i with (j + 1, ..., m) possibly left over the right edge. You can see the code below on how to update the dp table. This takes O(m^2) time per string so overall runtime is O(n^4) with a fairly low constant. We can get some more constant factor optimizations by doing some early breaks.
code
#ifdef LOCAL
#define _GLIBCXX_DEBUG
#endif
#include <bits/stdc++.h>
using namespace std;
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define fore(i, b, e) for (int i = (int)(b); i <= (int)(e); ++i)
#define ford(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define pb push_back
#define fi first
#define se second
#define all(x) (x).begin(), (x).end()
typedef vector<int> vi;
typedef pair<int, int> pii;
typedef long long i64;
typedef unsigned long long u64;
typedef long double ld;
typedef long long ll;
const int maxn = 500;
int n, m;
string s;
string t;
char d[maxn][maxn];
void scan() {
cin >> s;
n = s.size();
}
bool check() {
memset(d, 0, sizeof d);
m = t.size();
if (s[0] != t[0]) return false;
d[1][1] = 1;
fore(i, 2, n) fore(j, 1, m) if (t[j-1] == s[i-1]) {
if (d[i-1][j-1]) d[i][j] = 1;
if (d[i-1][m]) d[i][j] = 1;
if (j == 1) fore(k, 1, m-1) if (d[i-1][k]) { d[i][j] = 1; break; }
}
return d[n][m];
}
void solve() {
set<string> res;
set<string> seen;
forn(i, n) forn(len, n-i+1) if (len) {
t = s.substr(i, len);
if (seen.count(t)) continue;
seen.insert(t);
if (check()) res.insert(t);
}
for (auto s: res) {
cout << s << "\n";
}
}
int main() {
#ifdef LOCAL
freopen("e.in", "r", stdin);
#endif
scan();
solve();
#ifdef LOCAL
cerr << "Time elapsed: " << clock() / 1000 << " ms" << endl;
#endif
return 0;
}
O(n^3 log n)
For an O(n3·logn) solution, consider all occurences of t in s (these can be found in O(|s| + |t|) time). Then, consider extending a region as far as we can left by prepending a prefix of t, and extending a region as far as we can to the right by appending a suffix of t. This can be done with a stack or segment trees. Then, we have a set of intervals, and we want to check it covers 1, ..., n. This can also be done in time after sorting. Note, there is some tricky case in that if there can be gaps between the intervals, but these gaps need to consist of substrings of t. This can also be checked in linear time using suffix automaton.
code
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.io.BufferedWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.util.Collections;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main2 {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
StampStampStamp solver = new StampStampStamp();
solver.solve(1, in, out);
out.close();
}
static class StampStampStamp {
String s;
SegmentTreeFast st;
int[] r1;
int[] r2;
int[] left;
int[] right;
Pair<Integer, Integer>[] pp;
public void solve(int testNumber, InputReader in, OutputWriter out) {
s = in.next();
int n = s.length();
r1 = new int[n];
r2 = new int[n];
left = new int[n];
right = new int[n];
st = new SegmentTreeFast(n);
pp = new Pair[n];
HashSet<String> ss = new HashSet<>();
HashSet<String> t = new HashSet<>();
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
String test = s.substring(i, j + 1);
if (t.add(test) && ok(test, s)) {
ss.add(test);
}
}
}
ArrayList<String> ans = new ArrayList<>(ss);
Collections.sort(ans);
for (String x : ans) out.println(x);
}
boolean ok(String sub, String all) {
if (sub.charAt(0) != all.charAt(0) || sub.charAt(sub.length() - 1) != all.charAt(all.length() - 1))
return false;
HashSet<Character> tt = new HashSet<>();
for (char k : sub.toCharArray()) tt.add(k);
for (char k : all.toCharArray()) if (!tt.contains(k)) return false;
char[] c1 = (sub + "*" + all).toCharArray();
int[] z1 = ZAlgorithm.zAlgorithm(c1);
char[] c2 = (new StringBuffer(sub).reverse().toString() + "*" +
new StringBuffer(all).reverse().toString()).toCharArray();
int[] z2 = ZAlgorithm.zAlgorithm(c2);
int m = sub.length();
int n = all.length();
st.clear();
for (int i = n - 1; i >= 0; i--) {
int z = z1[i + m + 1];
if (z == 0) {
r1[i] = -1;
} else {
r1[i] = Math.max(i + z - 1, st.query(i + 1, i + z));
}
st.modify(i, i, r1[i]);
pp[i] = new Pair(-r1[i], i);
}
Arrays.sort(pp);
int idx = 0;
int min = n;
for (int i = n - 1; i >= 0; i--) {
while (idx < n && -pp[idx].u >= i) {
min = Math.min(min, pp[idx].v);
idx++;
}
left[i] = Math.min(min, i + 1);
}
st.clear();
int[] right = new int[n];
for (int i = 0; i < n; i++) {
int z = z2[n - i - 1 + m + 1];
if (z == 0) {
r2[i] = n + 1;
} else {
r2[i] = Math.min(i - z + 1, -st.query(i - z, i - 1));
}
st.modify(i, i, -r2[i]);
pp[i] = new Pair(r2[i], i);
}
Arrays.sort(pp);
idx = 0;
int max = -1;
for (int i = 0; i < n; i++) {
while (idx < n && pp[idx].u <= i) {
max = Math.max(max, pp[idx].v);
idx++;
}
right[i] = Math.max(i - 1, max);
}
ArrayList<Integer> evts = new ArrayList<>();
for (int i = 0; i < n; i++) {
if (z1[i + m + 1] == m) {
evts.add((i == 0 ? 0 : left[i - 1]) * 2 + 0);
evts.add(right[i] * 2 + 1);
}
}
Collections.sort(evts);
SuffixAutomaton.State[] automaton = SuffixAutomaton.buildSuffixAutomaton(sub);
int psum = 0;
for (int i = 0; i + 1 < evts.size(); i++) {
int k = evts.get(i);
if (k % 2 == 0) psum++;
else psum--;
if (psum == 0) {
int cur = k / 2 + 1;
int nxt = evts.get(i + 1) / 2;
int node = 0;
for (int j = cur; j < nxt; j++) {
int next = automaton[node].next[all.charAt(j)];
if (next == -1) return false;
node = next;
}
}
}
return evts.get(0) / 2 == 0 && evts.get(evts.size() - 1) / 2 == n - 1;
}
}
static class ZAlgorithm {
public static int[] zAlgorithm(char[] let) {
int N = let.length;
int[] z = new int[N];
int L = 0, R = 0;
for (int i = 1; i < N; i++) {
if (i > R) {
L = R = i;
while (R < N && let[R - L] == let[R])
R++;
z[i] = R - L;
R--;
} else {
int k = i - L;
if (z[k] < R - i + 1)
z[i] = z[k];
else {
L = i;
while (R < N && let[R - L] == let[R])
R++;
z[i] = R - L;
R--;
}
}
}
z[0] = N;
return z;
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
}
static class SuffixAutomaton {
public static SuffixAutomaton.State[] buildSuffixAutomaton(CharSequence s) {
int n = s.length();
SuffixAutomaton.State[] st = new SuffixAutomaton.State[Math.max(2, 2 * n - 1)];
st[0] = new SuffixAutomaton.State();
st[0].suffLink = -1;
int last = 0;
int size = 1;
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
int cur = size++;
st[cur] = new SuffixAutomaton.State();
st[cur].length = i + 1;
st[cur].firstPos = i;
int p = last;
for (; p != -1 && st[p].next[c] == -1; p = st[p].suffLink) {
st[p].next[c] = cur;
}
if (p == -1) {
st[cur].suffLink = 0;
} else {
int q = st[p].next[c];
if (st[p].length + 1 == st[q].length) {
st[cur].suffLink = q;
} else {
int clone = size++;
st[clone] = new SuffixAutomaton.State();
st[clone].length = st[p].length + 1;
System.arraycopy(st[q].next, 0, st[clone].next, 0, st[q].next.length);
st[clone].suffLink = st[q].suffLink;
for (; p != -1 && st[p].next[c] == q; p = st[p].suffLink) {
st[p].next[c] = clone;
}
st[q].suffLink = clone;
st[cur].suffLink = clone;
}
}
last = cur;
}
for (int i = 1; i < size; i++) {
st[st[i].suffLink].invSuffLinks.add(i);
}
return Arrays.copyOf(st, size);
}
public static class State {
public int length;
public int suffLink;
public List<Integer> invSuffLinks = new ArrayList<>(0);
public int firstPos = -1;
public int[] next = new int[128];
{
Arrays.fill(next, -1);
}
}
}
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1 << 16];
private int curChar;
private int numChars;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (this.numChars == -1) {
throw new InputMismatchException();
} else {
if (this.curChar >= this.numChars) {
this.curChar = 0;
try {
this.numChars = this.stream.read(this.buf);
} catch (IOException var2) {
throw new InputMismatchException();
}
if (this.numChars <= 0) {
return -1;
}
}
return this.buf[this.curChar++];
}
}
public String next() {
int c;
while (isSpaceChar(c = this.read())) {
;
}
StringBuilder result = new StringBuilder();
result.appendCodePoint(c);
while (!isSpaceChar(c = this.read())) {
result.appendCodePoint(c);
}
return result.toString();
}
public static boolean isSpaceChar(int c) {
return c == 32 || c == 10 || c == 13 || c == 9 || c == -1;
}
}
static class SegmentTreeFast {
int[] value;
int[] delta;
public SegmentTreeFast(int n) {
value = new int[2 * n];
for (int i = 0; i < n; i++) {
value[i + n] = getInitValue();
}
for (int i = 2 * n - 1; i > 1; i -= 2) {
value[i >> 1] = queryOperation(value[i], value[i ^ 1]);
}
delta = new int[2 * n];
Arrays.fill(delta, getNeutralDelta());
}
public void clear() {
int n = value.length / 2;
for (int i = 0; i < n; i++) {
value[i + n] = getInitValue();
}
for (int i = 2 * n - 1; i > 1; i -= 2) {
value[i >> 1] = queryOperation(value[i], value[i ^ 1]);
}
Arrays.fill(delta, getNeutralDelta());
}
int modifyOperation(int x, int y) {
return Math.max(x, y);
}
int queryOperation(int leftValue, int rightValue) {
return Math.max(leftValue, rightValue);
}
int deltaEffectOnSegment(int delta, int segmentLength) {
if (delta == getNeutralDelta()) return getNeutralDelta();
// Here you must write a fast equivalent of following slow code:
// int result = delta;
// for (int i = 1; i < segmentLength; i++) result = queryOperation(result, delta);
// return result;
return delta;
// return delta * segmentLength;
}
int getNeutralDelta() {
return -(1 << 29);
}
int getInitValue() {
return -(1 << 29);
}
int joinValueWithDelta(int value, int delta) {
if (delta == getNeutralDelta()) return value;
return modifyOperation(value, delta);
}
int joinDeltas(int delta1, int delta2) {
if (delta1 == getNeutralDelta()) return delta2;
if (delta2 == getNeutralDelta()) return delta1;
return modifyOperation(delta1, delta2);
}
void pushDelta(int i) {
int d = 0;
for (; (i >> d) > 0; d++) {
}
for (d -= 2; d >= 0; d--) {
int x = i >> d;
value[x >> 1] = joinNodeValueWithDelta(x >> 1, 1 << (d + 1));
delta[x] = joinDeltas(delta[x], delta[x >> 1]);
delta[x ^ 1] = joinDeltas(delta[x ^ 1], delta[x >> 1]);
delta[x >> 1] = getNeutralDelta();
}
}
int joinNodeValueWithDelta(int i, int len) {
return joinValueWithDelta(value[i], deltaEffectOnSegment(delta[i], len));
}
public int query(int from, int to) {
to = Math.min(to, value.length / 2 - 1);
from = Math.max(from, 0);
if (to < from) return getInitValue();
from += value.length >> 1;
to += value.length >> 1;
pushDelta(from);
pushDelta(to);
int res = 0;
boolean found = false;
for (int len = 1; from <= to; from = (from + 1) >> 1, to = (to - 1) >> 1, len <<= 1) {
if ((from & 1) != 0) {
res = found ? queryOperation(res, joinNodeValueWithDelta(from, len)) : joinNodeValueWithDelta(from, len);
found = true;
}
if ((to & 1) == 0) {
res = found ? queryOperation(res, joinNodeValueWithDelta(to, len)) : joinNodeValueWithDelta(to, len);
found = true;
}
}
if (!found) throw new RuntimeException();
return res;
}
public void modify(int from, int to, int delta) {
if (from > to) return;
from += value.length >> 1;
to += value.length >> 1;
pushDelta(from);
pushDelta(to);
int a = from;
int b = to;
for (; from <= to; from = (from + 1) >> 1, to = (to - 1) >> 1) {
if ((from & 1) != 0) {
this.delta[from] = joinDeltas(this.delta[from], delta);
}
if ((to & 1) == 0) {
this.delta[to] = joinDeltas(this.delta[to], delta);
}
}
for (int i = a, len = 1; i > 1; i >>= 1, len <<= 1) {
value[i >> 1] = queryOperation(joinNodeValueWithDelta(i, len), joinNodeValueWithDelta(i ^ 1, len));
}
for (int i = b, len = 1; i > 1; i >>= 1, len <<= 1) {
value[i >> 1] = queryOperation(joinNodeValueWithDelta(i, len), joinNodeValueWithDelta(i ^ 1, len));
}
}
}
static class Pair<U extends Comparable<U>, V extends Comparable<V>> implements Comparable<Pair<U, V>> {
public final U u;
public final V v;
public Pair(U u, V v) {
this.u = u;
this.v = v;
}
public int hashCode() {
return (u == null ? 0 : u.hashCode() * 31) + (v == null ? 0 : v.hashCode());
}
public boolean equals(Object o) {
if (this == o)
return true;
if (o == null || getClass() != o.getClass())
return false;
Pair<U, V> p = (Pair<U, V>) o;
return (u == null ? p.u == null : u.equals(p.u)) && (v == null ? p.v == null : v.equals(p.v));
}
public int compareTo(Pair<U, V> b) {
int cmpU = u.compareTo(b.u);
return cmpU != 0 ? cmpU : v.compareTo(b.v);
}
public String toString() {
return String.format("[Pair = (%s, %s)", u.toString(), v.toString());
}
}
}
Increasing Sequence
Be careful, it is not true in general that the set of valid k forms an interval. For example, consider this case [6, 4, 4, 6]. k = 7 works, by turning the array into [1, 3, 4, 6]. k = 9 works by turning the array into [3, 4, 5, 6]. But, k = 8 doesn't work since the middle two elements will stay the same.
Our solution will eliminate intervals of k that are bad if we choose to flip or not flip certain elements. Let F(i) be the set of intervals that is bad if we flip element i, and let T(i) be the set of intervals that is bad if we don't flip element i.
For each individual element, if we choose to flip it, and this adds the inequality xi ≤ k, so this adds the bad interval [0, xi - 1] to F(i).
For a pair of adjacent indices xi, xi + 1, we can check we can check what inequalities it imposes on k if we decide to flip or not flip certain elements (you can see the code for the exact cases).
After getting these bad intervals, it suffices to try to find some point such that this point is not in F(i) or not in T(i) for all i. This can be done with a line sweep.
code
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <cassert>
using namespace std;
const int N = 200500;
int B[N];
int W[N];
int bad = 0;
int getbad(int i) {
return (W[2 * i] > 0 && W[2 * i + 1] > 0) ? 1 : 0;
}
void add(int i, int w) {
bad -= getbad(i / 2);
W[i] += w;
bad += getbad(i / 2);
}
struct ev {
int t;
int i;
int d;
friend bool operator <(const ev& a, const ev& b) {
return a.t < b.t;
}
};
vector<ev> evts;
const int INF = 2e9 + 42;
void add(int l, int r, int i) {
evts.emplace_back(ev{l, i, 1});
evts.emplace_back(ev{r + 1, i, -1});
}
int A[N];
void die() {
puts("-1");
exit(0);
}
int T(int x) {
return 2 * x + 1;
}
int F(int x) {
return 2 * x;
}
int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d", &A[i]);
}
for (int i = 0; i < n; i++) {
add(0, A[i] - 1, F(i));
}
bool f = false;
for (int i = 0; i < n - 1; i++) {
int a = A[i];
int b = A[i + 1];
int s = a + b;
if (a == b) {
add(0, s, F(i + 1));
add(0, s, T(i));
add(s, INF, F(i));
add(s, INF, T(i + 1));
} else if (a > b) {
add(0, s, T(i));
add(s, INF, T(i + 1));
} else {
add(0, s, F(i + 1));
add(s, INF, F(i));
}
}
add(0, -1, 0); // touch
sort(evts.begin(), evts.end());
int goodk = -1;
for (int l = 0, r = 0; l < (int)evts.size(); l = r) {
if (evts[l].t >= INF - 2) {
break;
}
while (r < (int)evts.size() && evts[r].t == evts[l].t) {
r++;
}
for (int i = l; i < r; i++) {
add(evts[i].i, evts[i].d);
}
if (bad == 0) {
goodk = evts[l].t;
break;
}
}
printf("%d\n", goodk);
if (goodk != -1) {
assert(W[2 * n] == 0 || W[2 * n + 1] == 0);
for (int i = 0; i < n; i++) {
assert(W[2 * i] == 0 || W[2 * i + 1] == 0);
printf("%d ", (B[i] = ((W[2 * i] > 0) ? A[i] : goodk - A[i])));
}
printf("\n");
for (int i = 0; i < n - 1; i++) {
assert(B[i] < B[i + 1]);
}
for (int i = 0; i < n; i++) {
assert(B[i] >= 0);
}
}
}
Yet Another Binary Matrix
There may be some exponential time solutions, which is why bounds are set a bit low, but you can also solve this in polynomial time with the matroid intersection algorithm in O(n^4). It might also be possible to solve this with some hill climbing algorithms, the idea is to try to maximize rank(W(A, B + X)) + rank(W(R - A, C - B - X)). I'll describe the matroid intersection algorithm.
The matroids here in this case are M1 = set of columns X such W(A, X) is linearly independent, and M2 = set of columns Y such that the rank of W(R - A, C - Y) is at least |R| - |A| (here, the rank is the maximum size of a subset of rows that is linearly independent).
The algorithm is similar to bipartite matching (in fact, bipartite matching is a special case of the matroid intersection algorithm). The rough steps are as follows. Let S be some set of elements that is in 
. Initially S is empty. To increase the size of S, make a directed graph as follows:
- Make a node for every column in the graph. Also make a node for a "source" and a "sink".
- For simplicity, I'll refer to a node not in S as x and a node in S as y.
- For two nodes x, y, draw a directed edge from x to y if S + x - y belongs in M1.
- For two nodes x, y, draw a directed edge from x to y if S + x - y belongs in M2.
- Connect the source to x if S + x belongs in M1.
- Connect x to the sink if S + x belongs in M2.
We then find an "augmenting" path from the source to the sink, and we can xor this with our original set to get a set with one size larger (i.e. if it was in S remove it, otherwise add it).
The proof isn't too easy though, but you can google some papers about the algorithm for proof/more details (here and here).
Of course, this only deals with the unweighted case, so to add back B into consideration, we can give columns in B weight 1 and other columns with weight 0. We can adapt the algorithm above to find the maximum weight instead, and check it contains all elements of B.
code
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.io.BufferedWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.util.TreeSet;
import java.util.ArrayList;
import java.util.List;
import java.util.stream.Stream;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.util.Comparator;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
YetAnotherBinaryMatrix solver = new YetAnotherBinaryMatrix();
solver.solve(1, in, out);
out.close();
}
static class YetAnotherBinaryMatrix {
int n;
int na;
int nb;
int[][] mat;
int[] a;
int[] b;
long[] A;
long[] nA;
boolean[] ba;
boolean[] bb;
public void solve(int testNumber, InputReader in, OutputWriter out) {
n = in.nextInt();
na = in.nextInt();
nb = in.nextInt();
a = in.readIntArray(na);
b = in.readIntArray(nb);
ba = new boolean[n];
for (int x : a) ba[x - 1] = true;
bb = new boolean[n];
for (int x : b) bb[x - 1] = true;
mat = new int[n][n];
for (int i = 0; i < n; i++) {
char[] c = in.next().toCharArray();
for (int j = 0; j < n; j++)
mat[i][j] = c[j] - '0';
}
A = new long[n];
nA = new long[n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (ba[j]) A[i] = A[i] * 2 + mat[j][i];
else nA[i] = nA[i] * 2 + mat[j][i];
}
}
Basis bs = new Basis();
for (int i = 0; i < n; i++) bs.add(nA[i]);
if (bs.size < n - na) {
out.println(-1);
return;
}
boolean[] ss = new boolean[n];
while (true) {
int[] x = findPath(ss);
if (x == null) break;
for (int j : x) {
ss[j] = !ss[j];
}
}
int count = 0;
for (boolean w : ss) if (w) count++;
if (count != na) {
out.println(-1);
return;
}
count = 0;
for (int i = 0; i < n; i++) {
if (bb[i] && !ss[i]) {
out.println(-1);
return;
} else if (ss[i] && !bb[i]) {
count++;
}
}
out.println(count);
int[] ret = new int[count];
int idx = 0;
for (int i = 0; i < n; i++) if (ss[i] && !bb[i]) ret[idx++] = i + 1;
out.println(ret);
}
public boolean ok1(boolean[] ss) {
Basis bs = new Basis();
for (int j = 0; j < n; j++) {
if (ss[j]) {
if (!bs.add(A[j]))
return false;
}
}
return true;
}
public boolean ok2(boolean[] ss) {
Basis bs = new Basis();
for (int j = 0; j < n; j++) {
if (!ss[j]) {
bs.add(nA[j]);
}
}
return bs.size >= n - na;
}
public int[] findPath(boolean[] ss) {
int nnodes = n + 2;
List<Integer>[] graph = Stream.generate(ArrayList::new).limit(nnodes).toArray(List[]::new);
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (!ss[i] && ss[j]) {
ss[i] = true;
ss[j] = false;
if (ok1(ss)) graph[i].add(j);
if (ok2(ss)) graph[j].add(i);
ss[i] = false;
ss[j] = true;
}
}
}
for (int i = 0; i < n; i++) {
if (!ss[i]) {
ss[i] = true;
if (ok2(ss)) graph[nnodes - 1].add(i);
if (ok1(ss)) graph[i].add(nnodes - 2);
ss[i] = false;
}
}
int[] prev = new int[nnodes];
int[] dist = new int[nnodes];
Arrays.fill(prev, -1);
Arrays.fill(dist, 1 << 25);
TreeSet<Integer> ts = new TreeSet<>(Comparator.comparingInt(x -> dist[x] * (n + 10) + x));
dist[nnodes - 1] = 0;
ts.add(nnodes - 1);
while (ts.size() > 0) {
int f = ts.pollFirst();
for (int nxt : graph[f]) {
int weight = (nxt < n ? 100 * (bb[nxt] ? -1 : 1) * (ss[nxt] ? -1 : 1) : 0) + 1;
if (dist[f] + weight < dist[nxt]) {
ts.remove(nxt);
dist[nxt] = dist[f] + weight;
prev[nxt] = f;
ts.add(nxt);
}
}
}
if (prev[nnodes - 2] == -1) return null;
int[] p = new int[nnodes];
int cur = prev[nnodes - 2];
int id = 0;
do {
p[id++] = cur;
cur = prev[cur];
} while (cur != nnodes - 1);
return Arrays.copyOf(p, id);
}
}
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1 << 16];
private int curChar;
private int numChars;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int[] readIntArray(int tokens) {
int[] ret = new int[tokens];
for (int i = 0; i < tokens; i++) {
ret[i] = nextInt();
}
return ret;
}
public int read() {
if (this.numChars == -1) {
throw new InputMismatchException();
} else {
if (this.curChar >= this.numChars) {
this.curChar = 0;
try {
this.numChars = this.stream.read(this.buf);
} catch (IOException var2) {
throw new InputMismatchException();
}
if (this.numChars <= 0) {
return -1;
}
}
return this.buf[this.curChar++];
}
}
public int nextInt() {
int c;
for (c = this.read(); isSpaceChar(c); c = this.read()) {
;
}
byte sgn = 1;
if (c == 45) {
sgn = -1;
c = this.read();
}
int res = 0;
while (c >= 48 && c <= 57) {
res *= 10;
res += c - 48;
c = this.read();
if (isSpaceChar(c)) {
return res * sgn;
}
}
throw new InputMismatchException();
}
public String next() {
int c;
while (isSpaceChar(c = this.read())) {
;
}
StringBuilder result = new StringBuilder();
result.appendCodePoint(c);
while (!isSpaceChar(c = this.read())) {
result.appendCodePoint(c);
}
return result.toString();
}
public static boolean isSpaceChar(int c) {
return c == 32 || c == 10 || c == 13 || c == 9 || c == -1;
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(int[] array) {
for (int i = 0; i < array.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(array[i]);
}
}
public void println(int[] array) {
print(array);
writer.println();
}
public void close() {
writer.close();
}
public void println(int i) {
writer.println(i);
}
}
static class Basis {
public int size;
public long[] basis;
public Basis() {
basis = new long[100];
size = 0;
}
public boolean add(long x) {
for (int i = size - 1; i >= 0; i--) {
x = Math.min(x, x ^ basis[i]);
}
if (x == 0) return false;
basis[size++] = x;
for (int i = size - 2; i >= 0; i--) {
if (basis[i] > basis[i + 1]) {
long t = basis[i + 1];
basis[i + 1] = basis[i];
basis[i] = t;
}
}
return true;
}
}
}
