AcWing 1233. 全球变暖https://www.acwing.com/file_system/file/content/whole/index/content/4419714/
这洛谷上的题其实是上一道全球变暖问题的基础版，其实都是bfs对于联通块的问题，但是，洛谷这道就是只是在求连通块的数量
原题链接：https://www.luogu.com.cn/problem/P1451

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>

#define x first
#define y second
using namespace std;
typedef pair<int, int> PII;
const int N = 110;
bool st[N][N];
char g[N][N];
int n, m;
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, -1, 0, 1};
int cnt, res;
void bfs(int x, int y)
{   
    queue<PII> q;
    q.push({x, y});
    st[x][y] = true;

    while(q.size())  
    {
        auto t = q.front();
        q.pop();
        for(int i = 0; i < 4; i ++)
        {
            int a = t.x + dx[i], b = t.y + dy[i];
            if(a < 0 || a >= n || b < 0 || b >= m) continue;
            if(g[a][b] == '0') 
            {
                continue;
            }
            if(st[a][b]) continue;
            q.push({a,b});
            st[a][b] = true;
        }
    }
     cnt ++; 
}

int main()
{
    cin >> n >> m;
    for(int i = 0; i < n; i ++) cin >> g[i];

    for(int i = 0; i < n; i ++)
        for(int j = 0; j < m; j ++)
        if(!st[i][j] && g[i][j] != '0')
        {

            bfs(i, j); 

        }
        cout << cnt;
    return 0;
}