POJ-2536 Gopher II 二分图匹配

题目给定N个老鼠，M个洞，在规定的时间后，将有天敌来猎捉它们，每个老鼠都有一个速度，求一种躲藏的策略，使得受威胁的老鼠最少。（每个洞中只能藏一只老鼠）

该题可以将老鼠到达可及洞的情况看作是一个匹配，于是该题就是求一个老鼠到洞的最大匹配。

代码如下：

#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <cmath>
#define MAXN 100
using namespace std;

int n, m, s, v;
int G[MAXN+5][MAXN+5], visit[MAXN+5], marry[MAXN+5];

struct info
{
    double x, y, dist;
}d[MAXN+5];

double dist(double x1, double y1, double x2, double y2)
{
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}

int path(int u)
{
    for (int i = 1; i <= m; ++i) {
        if (!G[u][i] || visit[i]) {
            continue;
        }
        visit[i] = 1;
        if (!marry[i] || path(marry[i])) {
            marry[i] = u;
            return 1;
        }
    }
    return 0;
}

int main()
{
    double xx, yy; 
    int ans;
    while (scanf("%d %d %d %d", &n, &m, &s, &v) == 4) {
        ans = 0;
        memset(G, 0, sizeof (G));
        memset(marry, 0, sizeof (marry));
        for (int i = 1; i <= n; ++i) {
            scanf("%lf %lf", &d[i].x, &d[i].y);
        }
        for (int i = 1; i <= m; ++i) {
            scanf("%lf %lf", &xx, &yy);
            for (int j = 1; j <= n; ++j) {
                if (s*v >= dist(d[j].x, d[j].y, xx, yy))
                    G[j][i] = 1;  // j点到i洞可达
            }
        }
        for (int i = 1; i <= n; ++i) {
            memset(visit, 0, sizeof (visit));
            if (path(i)) {
                ++ans;
            }
        }
        printf("%d\n", n-ans);
    }
    return 0;
}