我发现…建二分图真的是门艺术。。这道题能想到用二分图真是太厉害了Orz

附上大佬博客：https://blog.csdn.net/zyy173533832/article/details/12654539

                         https://blog.csdn.net/lezg_bkbj/article/details/12260431

然后记得结果除2。。附上AC代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;

const int MAXN=605;
int n1,n2,k;
int g[MAXN][MAXN];
int link[MAXN];
bool vis[MAXN];

bool dfs(int u)
{
    for(int v=0;v<n2;v++)
        if(g[u][v]&&!vis[v])
        {
            vis[v]=true;
            if(link[v]==-1||dfs(link[v]))
            {
                link[v]=u;
                return true;
            }
        }
    return false;
}
int main()
{
    int n;
    char s[MAXN][MAXN];
    int id[MAXN][MAXN];
    int t;
    scanf("%d",&t);
    for(int tt=1;tt<=t;tt++)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%s",s[i]);
        int tol=0;
        memset(id,-1,sizeof(id));
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                if(s[i][j]=='#')
                {
                    id[i][j]=tol++;
                }
        memset(g,0,sizeof(g));
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
            {
                if(s[i][j]=='.')
                    continue;
                if(i+1<n&&s[i+1][j]=='#')
                    g[id[i][j]][id[i+1][j]]=1;
                if(j+1<n&&s[i][j+1]=='#')
                    g[id[i][j]][id[i][j+1]]=1;
                if(i>0&&s[i-1][j]=='#')
                    g[id[i][j]][id[i-1][j]]=1;
                if(j>0&&s[i][j-1]=='#')
                    g[id[i][j]][id[i][j-1]]=1;
            }
        int ans=0;
        n1=n2=tol;
        memset(link,-1,sizeof(link));
        for(int u=0;u<n1;u++)
        {
            memset(vis,0,sizeof(vis));
            if(dfs(u))
                ans++;
        }
        printf("Case %d: %d\n",tt,ans/2);//结果除2
    }
    return 0;
}