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BFSwhile(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BFSwhile(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BFSint 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;}
};