POJ - 3041 - Asteroids

题目来源:http://poj.org/problem?id=3041

匹配的相关概念及定理见:https://blog.csdn.net/moon_sky1999/article/details/81331795

可以等价为二分图中求最小顶点覆盖,由相关定理可得,最小顶点覆盖等于最大匹配,因此只需用匈牙利算法求出最大匹配即可。

代码:

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
int n,k,from[555];
bool vis[555];
vector <int> v[555];
bool match(int x) {
    for (int i = 0; i < v[x].size(); ++i) {
        int u=v[x][i];
        if (vis[u] == 0) {
            vis[u] = 1;
            if (from[u] == -1 || match(from[u])) {
                from[u] = x;
                return 1;
            }
        }
    }
    return 0;
}
int hungry() {
    int tot = 0;
    memset(from, -1, sizeof(from));
    for (int i = 1; i <= n; ++i) {
        memset(vis, 0, sizeof(vis));
        if (match(i))
            ++tot;
    }
    return tot;
}
int main() {
    while (~scanf("%d%d", &n, &k)) {
        for (int i = 1; i <= n; ++i)
            v[i].clear();
        int x, y;
        for (int i = 1; i <= k; ++i) {
            scanf("%d%d", &x, &y);
            v[x].push_back(y);
        }
        int cnt = hungry();
        printf("%d\n", cnt);
    }
    return 0;
}

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转载自blog.csdn.net/moon_sky1999/article/details/81332585