MarioYC's blog

By MarioYC, 13 years ago, In English

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I've been solving some problem about DFS applications, this is the first one I do working with directed graphs. First I established that the following conditions should be enough for a graph to be a cactus, after doing a DFS and getting the DFS tree:

  • DFS tree should contain all node
  • There are no cross edges or forward edges
  • No edge is a bridge, so every edge in the tree, has a back edge covering it
  • Back edges don't intersect (no edge is covered by two back edges) and they don't overlap.
  • Every node with two or more sons (except the root) should have a back edge starting in it.
  • Previous conditions are enough for the graph to be Strongly connected.

However I'm getting WA, could it be an implementation mistake? or some error in the conditions I've assumed?

#include <cstdio>
#include <cstring>
#include <vector>

using namespace std;

#define MAXN 10000
#define MAXE 10000

int E,to[MAXE],nxt[MAXE],last[MAXN];

void add_edge(int u, int v){
    to[E] = v; nxt[E] = last[u]; last[u] = E++;
}

int n,cont,parent[MAXN],level[MAXN],sons[MAXN],back[MAXN];
bool done[MAXN],forward,cross;

void dfs(int pos, int lvl){
    level[pos] = lvl;
    sons[pos] = 0;
    ++cont;
    
    for(int e = last[pos];e != -1;e = nxt[e]){
        int x = to[e];
        
        if(done[x]) cross = true;
        else if(parent[x] != -1){
            if(level[x] > level[pos]) forward = true;
            else back[pos] = (back[pos] == -1? level[x] : -2);
        }else{
            parent[x] = pos;
            ++sons[pos];
            
            dfs(x,lvl + 1);
        }
    }
    
    done[pos] = true;
}

bool is_cactus(){
    if(cont < n || forward || cross) return false;
    
    for(int i = 0;i < n;++i)
        if(back[i] == -2) return false;
    
    for(int i = 0;i < n;++i)
        if(i != 0 && sons[i] > 1 && back[i] == -1)
            return false;
    
    for(int i = 0;i < n;++i){
        if(sons[i] == 0){
            int e = level[i],pos = i;
            
            while(pos != 0){
                if(level[pos] == e){
                    if(back[pos] == -1) return false;
                    e = back[pos];
                }else if(back[pos] != -1) return false;
                
                pos = parent[pos];
            }
        }
    }
    
    return true;
}

int main(){
    int T,m;
    
    scanf("%d",&T);
    
    while(T--){
        scanf("%d %d",&n,&m);
        
        memset(last,-1,sizeof last);
        E = 0;
        
        for(int i = 0,u,v;i < m;++i){
            scanf("%d %d",&u,&v);
            add_edge(u,v);
        }
        
        memset(back,-1,sizeof back);
        memset(parent,-1,sizeof parent);
        memset(done,false,sizeof done);
        forward = cross = false;
        parent[0] = -2;
        cont = 0;
        
        dfs(0,0);
        
        puts(is_cactus()? "YES" : "NO");
    }
    
    return 0;
}

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13 years ago, # |
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It's a tricky problem, I can't seem to see any problem with your solution. My solution for this problem is very simple.

first you see if it is a SCC.

then you run 1 dfs, with vectors seen[], and finished[]. What you are looking for is if a the dfs reaches a finished node, that would mean that that some edge takes part in more than one cicle, and it is not a cactus. Because we know every node is in a cicle, and if we reach a finished node, we ca "re-use" part of the cicle from that node and make a new cicle using the same edge, it is not easy to convince yourself of this, but if you make a few drawings, you will see.

if you like I can post my code.