二分图最大匹配(匈牙利算法) POJ 3041 Asteroids

题目传送门

 /*
     题意：每次能消灭一行或一列的障碍物，要求最少的次数。
     匈牙利算法：把行和列看做两个集合，当有障碍物连接时连一条边，问题转换为最小点覆盖数==二分图最大匹配数
         趣味入门：http://blog.csdn.net/dark_scope/article/details/8880547
 */
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <vector>
 using namespace std;
 const int MAXN = 5e2 + ;
 const int INF = 0x3f3f3f3f;
 vector<int> G[MAXN];
 bool vis[MAXN];
 int lk[MAXN];
 bool DFS(int u)
 {
     for (int i=; i<G[u].size (); ++i)
     {
         int v = G[u][i];
         if (!vis[v])
         {
             vis[v] = true;
             if (lk[v] == - || DFS (lk[v]))
             {
                 lk[v] = u;    return true;
             }
         }
     }
     return false;
 }
 int hungary(int n)
 {
     int res = ;    memset (lk, -, sizeof (lk));
     for (int i=; i<=n; ++i)
     {
         memset (vis, false, sizeof (vis));
         if (DFS (i))    res++;
     }
     return res;
 }
 int main(void)        //POJ 3041 Asteroids
 {
 //    freopen ("POJ_3041.in", "r", stdin);
     int n, k;
     while (scanf ("%d%d", &n, &k) == )
     {
         for (int i=; i<=n; ++i)    G[i].clear ();
         for (int i=; i<=k; ++i)
         {
             int x, y;    scanf ("%d%d", &x, &y);
             G[x].push_back (y);
         }
         printf ("%d\n", hungary (n));
     }
     return ;
 }