The problem requires us to find the maximum possible sum of integers written on the whiteboards after performing a series of operations. Let's break down the problem and provide a step-by-step solution.

Observations:

Exactly n items out of a1, a2, ..., an, b1, b2, ..., bm will remain on the whiteboard at the end. The value bm will always remain on the board at the end.

Sort the list of integers ai in non-decreasing order. Add the value bm to the final sum. Iterate through the remaining (n + m — 1) integers: a. If the current integer is one of b1, b2, ..., bm, skip it. b. Otherwise, add it to the final sum.

The reason for adding bm initially is that it will always remain on the board, ensuring its contribution to the maximum sum. Then, we have the freedom to choose (n — 1) out of the remaining (n + m — 1) integers to add to the sum.

testCases = int(input())

for _ in range(testCases):
    n, m = map(int, input().split())
    A = list(map(int, input().split()))
    B = list(map(int, input().split()))
    arr = A + B[:-1]
    arr.sort(reverse=True)

    print(B[-1] + sum(arr[:n - 1]))

small code fix on A

Sorting the list of integers takes O((n + m) log(n + m)) time. The iteration through the remaining integers takes O(n + m) time. Therefore, the overall time complexity is O((n + m) log(n + m)).

Think about how you can use graph traversal algorithms, such as depth-first search (DFS) or breadth-first search (BFS), to find the connected components in the graph.

Consider using a visited array or set to keep track of visited characters during the traversal.

Remember to update the minimum cost within each connected component.

The problem requires us to find the minimum amount of gold Vova needs to spend to spread a rumor in the settlement of Overcity. The characters in Overcity are connected through friendships, and each character has a cost associated with spreading the rumor.

Approach:

1. Represent the friendships between characters as a graph.
2. Assign the cost of spreading the rumor to each character.
3. Find the connected components in the graph.
4. Calculate the sum of the minimum values in each connected component.

import collections

n, m = list(map(int, input().split())) nums = list(map(int, input().split()))

INF = 10 ** 9 + 13

arr = []

for index in range(n): arr.append([])

for index in range(m): v, u = list(map(int, input().split()))

v -= 1
u -= 1

arr[v].append(u)
arr[u].append(v)

used = [False] * n res = INF

def bfs(v): global arr, used, res

q = collections.deque([v])
used[v] = True

while q:
    u = q.pop()

    res = min(res, nums[u])

    for w in arr[u]:
        if not used[w]:
            q.append(w)
            used[w] = True

ans = 0

for index in range(n): if not used[index]: res = INF bfs(index) ans += res

print(ans)

O(nodes + edges)

Think about the initial selection of pairs. Which individuals should be chosen to maximize the number of talks?

Consider the case where the maximum sociability is greater than or equal to the sum of the remaining sociabilities. How does this affect the optimal solution?

Analyze the case where the maximum sociability is smaller than the sum of the remaining sociabilities. What is the maximum number of talks that can be achieved?

The problem requires us to determine the optimal arrangement of conversations between people in a meeting in order to maximize the total number of talks. Each person has a limited sociability, represented by a non-negative integer, and after a certain number of talks, they leave the meeting. The goal is to find the maximum number of talks possible.

Approach:

1. Identify the person with the highest sociability and pair them up with another person who also has a high sociability.
2. Decrease the sociability of the two paired individuals by 1.
3. Repeat steps 1 and 2 while there are at least two people with positive sociability remaining.
4. Calculate the total number of talks.

import heapq
def solve():
    n = int(input())
    a = list(map(int, input().split()))
    heap = []
    
    for i, soc in enumerate(a):
        if soc > 0:
            heapq.heappush(heap, (-soc, i + 1))
    
    talks = []
    while len(heap) > 1:
        top1, p1 = heapq.heappop(heap)
        top2, p2 = heapq.heappop(heap)
        talks.append(sorted([p1, p2]))
        
        top1 += 1
        top2 += 1
        
        if top1 < 0:
            heapq.heappush(heap, (top1, p1))
        
        if top2 < 0:
            heapq.heappush(heap, (top2, p2))
    

    print(len(talks))
    for p1, p2 in talks:
        print(p1, p2)
    
    
def main():
    t = int(input())
    for i in range(t):
        solve()
main()

If s is the total sum of all sociability's then it is O(slog n).

How do we detect cycles in an undirected graph?

The task involves determining if an undirected graph, where nodes represent cells and edges represent adjacent cells with the same color, contains any cycles.

Run dfs / bfs, if an edge leads you to a visited node, then there must be a cycle.

from collections import deque

def is_valid_cell(x, y, n, m):
    return 0 <= x < n and 0 <= y < m

def has_cycle(grid, n, m):
    # directions to move in the grid
    dx = [0, 0, 1, -1]
    dy = [1, -1, 0, 0]

    # visited set to track visited cells
    visited = set()

    # perform bfs from each cell
    for i in range(n):
        for j in range(m):
            if (i, j) in visited:
                continue

            color = grid[i][j]
            q = deque([(i, j, None)])

            while q:
                x, y, prev = q.popleft()
                if (x, y) in visited:
                    # cycle found
                    return True

                visited.add((x, y))

                for k in range(4):
                    nx, ny = x + dx[k], y + dy[k]

                    if not is_valid_cell(nx, ny, n, m) or grid[nx][ny] != color:
                        continue

                    if (nx, ny) != prev:
                        q.append((nx, ny, (x, y)))

    # no cycle found
    return False

# read input
n, m = map(int, input().split())
grid = [input() for _ in range(n)]

# check if there is a cycle
if has_cycle(grid, n, m):
    print("Yes")
else:
    print("No")

def is_valid(x, y, n, m):
    return x >= 0 and x < n and y >= 0 and y < m

def has_cycle(grid):
    visited = set()

    # direction vectors for neighbors
    dx = [1, -1, 0, 0]
    dy = [0, 0, 1, -1]

    for i in range(len(grid)):
        for j in range(len(grid[0])):
            if (i, j) not in visited:
                stack = [(i, j, None)]

                while stack:
                    x, y, parent = stack.pop()

                    visited.add((x, y))

                    for k in range(4):
                        nx, ny = x + dx[k], y + dy[k]

                        if is_valid(nx, ny, len(grid), len(grid[0])) and grid[nx][ny] == grid[x][y]:
                            if (nx, ny) not in visited:
                                stack.append((nx, ny, (x, y)))
                            elif (nx, ny) != parent:
                                return True

    return False

# read input
n, m = map(int, input().split())
grid = [input() for _ in range(n)]

# check if there is a cycle
if has_cycle(grid):
    print("Yes")
else:
    print("No")

Since we are visiting each node of a graph the time complexity is O(Rows * COLS).

From the input we are given that the total length of our path doesn’t exceed 10^5.

How can we find the shortest path in an undirected graph.

This problem involves finding the shortest path between two points on a chess field, given a set of allowed cells. We can model the chess field as an undirected graph and use breadth-first search (BFS) to find the shortest path between the two points.

To ensure that the king only moves along the allowed cells, we can represent the allowed cells as a set and only consider a path if all of its cells are in the set of allowed cells.

By using BFS, we guarantee that we find the shortest path between the two points, as BFS explores nodes in order of their distance from the starting node. If there is no path between the two points along the allowed cells, we can detect this by the fact that the BFS algorithm does not visit the ending node.

from collections import deque

def main():
    # Input
    x0, y0, x1, y1 = map(int, input().split())
    n = int(input())
    allowed = set()
    for i in range(n):
        r, a, b = map(int, input().split())
        for i in range(a, b + 1):
            allowed.add((r, i))

    # BFS
    queue = deque([(x0, y0)])
    visited = set([(x0, y0)])
    neighbors = [[-1, 0], [0, -1], [0, 1], [1, 0],
                 [-1, 1], [1, -1], [1, 1], [-1, -1]]
    level = 0
    while queue:
        level += 1
        for _ in range(len(queue)):
            x, y = queue.popleft()
            if (x, y) == (x1, y1):
                return level - 1
            for dx, dy in neighbors:
                nx, ny = x + dx, y + dy
                if (nx, ny) in allowed and (nx, ny) not in visited:
                    visited.add((nx, ny))
                    queue.append((nx, ny))

    # No path found
    return -1

print(main())

The time complexity is equal to the number of allowed paths. If the number of allowed paths is n, then it is O(n)

Try to simulate the given sequence of operations and maintain the state of the heap at each step.

Think about the conditions when an insert operation is required to make the sequence consistent.

One way to approach this problem is to simulate the given sequence of operations and maintain a set of all the elements currently present in the heap. Whenever a getMin operation is encountered, we check if the minimum element is present in the set. If not, we add an insert operation to insert the minimum element. Similarly, whenever a removeMin operation is encountered, we check if the set is empty. If it is, we add an insert operation to insert any element into the heap.

If the current operation is insert x, then add element x to the heap.

If current operation is removeMin, then if heap is not empty, then simply remove minimal element, otherwise if heap is empty, add operation insert x, where x can be any number, and then apply removeMin.

If current operation is getMin x then do follows:

1.While the heap is not empty and its minimal element is less than x, apply operation removeMin.

2.Now if the heap is empty or its minimal element is not equal to x, apply operation insert x.

3.Apply operation getMin x.

import heapq
def main():
    num_operations = int(input())
    
    output = []
    heap = []
    for i in range(num_operations):
        record = list(input().split())
    
        if record[0] == 'insert':
            heapq.heappush(heap, int(record[1]))
            output.append(record)
        
        elif record[0] == 'removeMin':
            if not heap:
                output.append(['insert', '1'])
                heapq.heappush(heap, 1)
            heapq.heappop(heap)
            output.append(record)
        
        else:
            x = int(record[1])
            
            while heap and heap[0] < x:
                heapq.heappop(heap)
                output.append(['removeMin'])
            
            if not heap or heap[0] > x:
                heapq.heappush(heap, x)
                output.append(['insert',  str(x)])
            
            output.append(record)
    
    print(len(output))
    for line in output:
        print(' '.join(line))

main()

O(N*log(N))