hdu 4185 二分图匹配

题意用1*2的木板覆盖矩阵中的‘#’，（木板要覆盖的只能是‘#’），问最多能用几个木板覆盖

将#抽象为二分图的点，一个木板就是一个匹配，注意最后结果要除以2

Sample Input

1
6
......
.##...
.##...
....#.
....##
......

Sample Output

Case 1: 3

 

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<cstring>
 #include<cmath>
 #include<queue>
 #include<map>
 using namespace std;
 #define MOD 1000000007
 const int INF=0x3f3f3f3f;
 const double eps=1e-;
 typedef long long ll;
 #define cl(a) memset(a,0,sizeof(a))
 #define ts printf("*****\n");
 const int MAXN=;
 int n,m,tt;
 char s[MAXN][MAXN];
 int uN,vN;//u,v的数目，使用前面必须赋值
 int g[MAXN][MAXN];//邻接矩阵
 int linker[MAXN];
 bool used[MAXN];
 int num[MAXN][MAXN];
 bool dfs(int u)
 {
     for(int v=;v<vN;v++)
     if(g[u][v]&&!used[v])
     {
         used[v]=true;
         if(linker[v]==-||dfs(linker[v]))
         {
             linker[v] = u;
             return true;
         }
     }
     return false;
 }
 int hungary()
 {
     int res=;
     memset(linker,-,sizeof(linker));
     for(int u=;u<uN;u++)
     {
         memset(used,false,sizeof(used));
         if(dfs(u))res++;
     }
     return res;
 }
 int main()
 {
     int i,j,k;
     #ifndef ONLINE_JUDGE
     freopen("1.in","r",stdin);
     #endif
     int ca=;
     scanf("%d",&tt);
     while(tt--)
     {
         scanf("%d",&n);
         cl(g);
         int tot=;
         cl(num);
         for(i=;i<n;i++)
         {
             scanf("%s",&s[i]);
             for(j=;j<n;j++)
             {
                 if(s[i][j]=='#')    num[i][j]=tot++;
             }
         }

         uN=vN=tot;
         for(i=;i<n;i++)
         {
             for(j=;j<n;j++)
             {
                 if(s[i][j]=='#')
                 {
                     if(s[i+][j]=='#'&&i+<n)  g[num[i][j]][num[i+][j]]=;
                     if(s[i][j+]=='#'&&j+<n)  g[num[i][j]][num[i][j+]]=;
                     if(s[i-][j]=='#'&&i->=)  g[num[i][j]][num[i-][j]]=;
                     if(s[i][j-]=='#'&&j->=)  g[num[i][j]][num[i][j-]]=;
                 }
             }
         }
         ca++;
         printf("Case %d: %d\n",ca,hungary()/);
     }
 }