BFS解决多源最短路相关leetcode算法题

1.01矩阵

01矩阵

class Solution {
    int dx[4] = {0,0,1,-1};
    int dy[4] = {1,-1,0,0};
public:
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        //正难则反，找0到1的最短距离
        int m = mat.size(),n = mat[0].size();
        queue<pair<int,int>> q;
        //通过此数组对应位置的值既可以标识某个位置是否已访问过，也可在此基础上往外扩展
        vector<vector<int>> dist(m,vector<int>(n,-1));
        //if(dist[][] == -1) 表示此位置没被访问过
        //if(dist[][] != -1) 表示此位置被访问过
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(mat[i][j] == 0)
                {
                    dist[i][j] = 0;
                    q.push({i,j});
                }
            }
        }
        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 && mat[x][y]&& dist[x][y]==-1)
                {
                    dist[x][y] = dist[a][b]+1;
                    q.push({x,y});
                }
            }
        }
        return dist;
    }
};

2.飞地的数量

飞地的数量

class Solution {
    int dx[4] = {0,0,1,-1};
    int dy[4] = {1,-1,0,0};
public:
    int numEnclaves(vector<vector<int>>& grid) {
        int m = grid.size(),n = grid[0].size();
        vector<vector<bool>> vis(m,vector<bool>(n));
        queue<pair<int,int>> q;
        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});
                        vis[i][j] = true;
                    }
                }
            }
        }
        //多源BFS
        while(q.size())
        {
            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&& !vis[x][y])
                {
                    q.push({x,y});
                    vis[x][y] = true;
                }
            }
        }
        //遍历统计
        int ret = 0;
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(grid[i][j]==1 && !vis[i][j]) ret++;
            }
        }

        return ret;
    }
};

3.地图中的最高点

地图中的最高点

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

4.地图分析

地图分析

class Solution {
    int dx[4] = {0,0,1,-1};
    int dy[4] = {1,-1,0,0};
public:
    int maxDistance(vector<vector<int>>& grid) {
        //正难则反
        int m=grid.size(),n=grid[0].size();
        vector<vector<int>> dist(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(grid[i][j])
                {
                    dist[i][j] =0;
                    q.push({i,j});
                }
            }
        }
        //多源BFS
        int ret = -1;
        while(q.size())
        {
            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 && dist[x][y] == -1)
                {
                    dist[x][y] = dist[a][b]+1;
                    q.push({x,y});
                    ret = max(ret,dist[x][y]);
                }
            }
        }
        return ret;
    }
};