POJ 3041 Asteroids

解题思路：将行列分别对应二部图的两个set(行，列)->（left， right)，问题转化为最小顶点覆盖问题，进而转化为求解最大流问题

#include <iostream>
using namespace std;
#define Num 505

bool edge[Num][Num];
int father[Num];
bool visit[Num];
int N;

int Search(int x)
{
    for (int i = 1; i <= N; i++)
    {
        if (!visit[i] && edge[x][i])
        {
            visit[i] = true;
            if (father[i] == 0 || Search(father[i]))
            {
                father[i] = x;
                return 1;
            }
        }
    }
    return 0;
}

int main()
{
    int K, R, C, t = 0;
    scanf("%d%d", &N, &K);
    for (int i = 0; i < K; i++)
    {
        scanf("%d%d", &R, &C);
        edge[R][C] = true;
    }
    for (int i = 1; i <= N; i++)
    {
        memset(visit, 0, sizeof(visit));
        t += Search(i);
    }
    printf("%d\n", t);
    return 0;
}