2060. 奶牛选美 - AcWing题库

思路：只有两个连通块，先标记一个连通块。之后将这个连通块全部放入队列跑一个bfs即可。

#include <bits/stdc++.h>
using namespace std;
const int N = 121;
int fx[] = {0, 0, 1, -1};
int fy[] = {1, -1, 0, 0};
bool vis[N][N];
string ph[N];
struct no {int x,y,cnt;
};
int main()
{int n,m; cin>>n>>m;for(int i = 0; i < n; ++i) cin>>ph[i];bool ok = false;for(int i = 0; i < n; ++i) {for(int j = 0; j < m; ++j) {if(ph[i][j] == 'X') {auto dfs = [&](auto &&self, int x, int y) -> void {vis[x][y] = 1;for(int i = 0; i < 4; ++i) {int xx = x + fx[i], yy = y + fy[i];if(xx < 0 || xx >= n || yy < 0 || yy >= m) continue;if(ph[xx][yy] == '.' || vis[xx][yy]) continue;self(self, xx ,yy);}};dfs(dfs,i,j);ok = true;break;}}if(ok) break;}queue<no> q;for(int i = 0; i < n; ++i) {for(int j = 0; j < m; ++j) {if(vis[i][j]) q.push({i, j, 0});}}while(q.size()) {auto tmp = q.front(); q.pop();for(int i = 0; i < 4; ++i) {int xx = tmp.x + fx[i], yy = tmp.y + fy[i];if(xx < 0 || xx >= n || yy < 0 || yy >= m) continue;if(vis[xx][yy]) continue;vis[xx][yy] = 1;if(ph[xx][yy] == 'X') {cout<<tmp.cnt<<endl;return 0;}q.push({xx, yy, tmp.cnt + 1});}}
}