hdu - 1068 Girls and Boys (二分图最大独立集+拆点)

http://acm.hdu.edu.cn/showproblem.php?pid=1068

因为没有指定性别,所以要拆点,把i拆分i和i’ 那么U=V-M (M是最大匹配,U最大独立集,V是顶点数)

2U=2V-2M  所以 U=n-M'/2. (没怎么看明白)  但是不这样会wa.

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <vector>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <string>
 #include <set>
 #include <functional>
 #include <numeric>
 #include <sstream>
 #include <stack>
 #include <map>
 #include <queue>
 #pragma comment(linker, "/STACK:102400000,102400000")
 #define CL(arr, val)    memset(arr, val, sizeof(arr))
 
 #define ll long long
 #define inf 0x7f7f7f7f
 #define lc l,m,rt<<1
 #define rc m + 1,r,rt<<1|1
 #define pi acos(-1.0)
 
 #define L(x)    (x) << 1
 #define R(x)    (x) << 1 | 1
 #define MID(l, r)   (l + r) >> 1
 #define Min(x, y)   (x) < (y) ? (x) : (y)
 #define Max(x, y)   (x) < (y) ? (y) : (x)
 #define E(x)        (1 << (x))
 #define iabs(x)     (x) < 0 ? -(x) : (x)
 #define OUT(x)  printf("%I64d\n", x)
 #define lowbit(x)   (x)&(-x)
 #define Read()  freopen("a.txt", "r", stdin)
 #define Write() freopen("b.txt", "w", stdout);
 #define maxn 1000000000
 #define N 510
 #define mod 1000000000
 using namespace std;
 
 vector<int>G[N];
 int link[N];
 bool vis[N];
 
 bool dfs(int u)
 {
     for(int i=;i<G[u].size();i++)
     {
         int v=G[u][i];
         if(!vis[v])
         {
             vis[v]=true;
             if(link[v]==-||dfs(link[v]))
             {
                 link[v]=u;
                 return true;
             }
         }
     }
     return false;
 }
 int main()
 {
     //Read();
     int n,a,b,c;
     while(~scanf("%d",&n))
     {
         for(int i=;i<n;i++) G[i].clear();
         for(int i=;i<n;i++)
         {
             scanf("%d: (%d)",&a,&b);
            // printf("%d %d\n",a,b);
             for(int i=;i<b;i++)
             {
                 scanf("%d",&c);
                 G[a].push_back(c);
             }
         }
         int ans=;
         memset(link,-,sizeof(link));
         for(int i=;i<n;i++)
         {
             memset(vis,,sizeof(vis));
             if(dfs(i)) ans++;
         }
         printf("%d\n",n-ans/);
     }
     return ;
 }