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【多源BFS】01 矩阵 / 飞地的数量 / 地图中的最高点 / 地图分析 / 腐烂的苹果

news 来源：原创 2025/5/10 13:51:23

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动图描述

01 矩阵

01 矩阵

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class Solution {
public:
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        int dx[4] = {1, -1, 0, 0}, dy[4] = {0, 0, 1, -1};
        int m = mat.size(), n = mat[0].size();
        vector<vector<int>> ret(m, vector<int>(n, -1));
        queue<pair<int, int>> q;
        for (int i = 0; i < m; i++)
            for (int j = 0; j < n; j++)
                if (mat[i][j] == 0)
                {
                    q.push({i, j});
                    ret[i][j] = 0;
                }
        while (q.size())
        {
            auto [a, b] = q.front();
            q.pop();
            for (int i = 0; i < 4; i++)
            {
                int x = a + dx[i], y = b + dy[i];
                if (x >= 0 && x < m && y >= 0 && y < n && ret[x][y] == -1)
                {
                    ret[x][y] = ret[a][b] + 1;
                    q.push({x, y});
                }
            }
        }
        return ret;
    }
};

飞地的数量

飞地的数量

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class Solution {
public:
    int numEnclaves(vector<vector<int>>& grid) {
        int dx[4] = {1, -1, 0, 0}, dy[4] = {0, 0, 1, -1};
        int m = grid.size(), n = grid[0].size();
        queue<pair<int, int>> q;
        bool used[501][501] = {};
        for (int i = 0; i < m; i++)
            for (int j = 0; j < n; j++)
                if (i == 0 || i == m - 1 || j == 0 || j == n - 1)
                    if (grid[i][j] == 1) 
                    {
                        q.push({i, j});
                        used[i][j] = true;
                    }
        while (q.size())
        {
            auto [a, b] = q.front();
            q.pop();
            for (int i = 0; i < 4; i++)
            {
                int x = a + dx[i], y = b + dy[i];
                if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] && !used[x][y])
                {
                    used[x][y] = true;
                    q.push({x, y});
                }
            }
        }
        int ret = 0;
        for (int i = 0; i < m; i++)
            for (int j = 0; j < n; j++)
                if (grid[i][j] && !used[i][j])
                    ret++;
        return ret;
    }
};

地图中的最高点

地图中的最高点

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class Solution {
public:
    vector<vector<int>> highestPeak(vector<vector<int>>& isWater) {
        int dx[4] = {1, -1, 0, 0}, dy[4] = {0, 0, 1, -1};
        int m = isWater.size(), n = isWater[0].size();
        queue<pair<int, int>> q;
        vector<vector<int>> ret(m, vector<int>(n, -1));
        for (int i = 0; i < m; i++)
            for (int j = 0; j < n; j++)
                if (isWater[i][j] == 1)
                {
                    q.push({i, j});
                    ret[i][j] = 0;
                }
        while (q.size())
        {
            auto [a, b] = q.front();
            q.pop();
            for (int i = 0; i < 4; i++)
            {
                int x = a + dx[i], y = b + dy[i];
                if (x >= 0 && x < m && y >= 0 && y < n && ret[x][y] == -1)
                {
                    ret[x][y] = ret[a][b] + 1;
                    q.push({x, y});
                }
            }
        }
        return ret;
    }
};

地图分析

地图分析

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class Solution {
public:
    int maxDistance(vector<vector<int>>& grid) {
        int dx[4] = {1, -1, 0, 0}, dy[4] = {0, 0, 1, -1};
        int n = grid.size();
        queue<pair<int, int>> q;
        bool used[101][101] = {};
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                if (grid[i][j] == 1)
                    q.push({i, j});
        if (q.size() == n * n || q.size() == 0) return -1;
        int ret = -1;
        while (q.size())
        {
            ret++;
            int sz = q.size();
            while (sz--)
            {
                auto [a, b] = q.front();
                q.pop();
                for (int i = 0; i < 4; i++)
                {
                    int x = a + dx[i], y = b + dy[i];
                    if (x >= 0 && x < n && y >= 0 && y < n && !grid[x][y] && !used[x][y])
                    {
                        used[x][y] = true;
                        q.push({x, y});
                    }
                }
            }
        }
        return ret;
    }
};

腐烂的苹果

腐烂的苹果

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class Solution {
public:
    int rotApple(vector<vector<int> >& grid) {
        int dx[4] = {1, -1, 0, 0}, dy[4] = {0, 0, 1, -1};
        int m = grid.size(), n = grid[0].size();
        queue<pair<int, int>> q;
        for (int i = 0; i < m; i++)
            for (int j = 0; j < n; j++)
                if (grid[i][j] == 2)
                    q.push({i, j});
        int min = -1;
        while (q.size())
        {
            min++;
            int sz = q.size();
            while (sz--)
            {
                auto [a, b] = q.front();
                q.pop();
                for (int k = 0; k < 4; k++)
                {
                    int x = a + dx[k], y = b + dy[k];
                    if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1)
                    {
                        grid[x][y] = 2;
                        q.push({x, y});
                    }
                }
            }
        }
        for (int i = 0; i < m; i++)
            for (int j = 0; j < n; j++)
                if (grid[i][j] == 1) 
                    return -1;
        return min;
    }
};