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 + 10;
const int INF = 0x3f3f3f3f;
vector<int> G[MAXN];
bool vis[MAXN];
int lk[MAXN];

bool DFS(int u)
{
    for (int i=0; i<G[u].size (); ++i)
    {
        int v = G[u][i];
        if (!vis[v])
        {
            vis[v] = true;
            if (lk[v] == -1 || DFS (lk[v]))
            {
                lk[v] = u;    return true;
            }
        }
    }
    return false;
}

int hungary(int n)
{
    int res = 0;    memset (lk, -1, sizeof (lk));
    for (int i=1; 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) == 2)
    {
        for (int i=1; i<=n; ++i)    G[i].clear ();
        for (int i=1; i<=k; ++i)
        {
            int x, y;    scanf ("%d%d", &x, &y);
            G[x].push_back (y);
        }

        printf ("%d
", hungary (n));
    }

    return 0;
}