poj 3041 二分图最小点集覆盖

定理：二分图的最大匹配=最小点覆盖。

思路：将所有行看做点集X，所有列看做点集Y，如果在[i, j]处有小行星，则建立一条从i到j的边，然后求最大匹配即为最小点覆盖数即为答案。

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

const int N = 501;
const int M = N * N;
int head[N];
int mark[N];
bool visit[N];
int n, m, e;

void init()
{
    e = 0;
    memset( head, -1, sizeof(head) );
}

struct Edge
{
    int v, next;
} edge[M];

void addEdge( int u, int v )
{
    edge[e].v = v;
    edge[e].next = head[u];
    head[u] = e++;
}

int dfs( int u )
{
    for ( int i = head[u]; i != -1; i = edge[i].next )
    {
        int v = edge[i].v;
        if ( !visit[v] )
        {
            visit[v] = 1;
            if ( mark[v] == -1 || dfs( mark[v] ) )
            {
                mark[v] = u;
                return 1;
            }
        }
    }
    return 0;
}

int hungary()
{
    memset( mark, -1, sizeof(mark) );
    int res = 0;
    for ( int i = 1; i <= n; i++ )
    {
        memset( visit, 0, sizeof(visit) );
        res += dfs(i);
    }
    return res;
}

int main ()
{
    while ( scanf("%d%d", &n, &m) != EOF )
    {
        init();
        while ( m-- )
        {
            int u, v;
            scanf("%d%d", &u, &v);
            addEdge( u, v );
        }
        int ans = hungary();
        printf("%d
", ans);
    }
    return 0;
}