poj 2532 Gopher II（最大匹配）

题意：有n个晏鼠 m个洞，给出他们的坐标，每个晏鼠的速度都是一样的V 如果在s秒内，它们没有跑到洞是就会被老鹰吃掉，每个洞只能容纳一个晏鼠 求最后有多少晏鼠会被吃掉

思路： 这是个比较简单的二分图最大匹配问题，建图也相对简单，只在晏鼠到洞的距离小于V*S 就把他们连起来 求出最大匹配 再用n-最大匹配

#include <stdio.h>
#include <string.h>
#include <math.h>
#define M 110
int n,m;
int map[M][M],link[M],vis[M];
struct node
{
    double x,y;
}g[M],h[M];
double len (node a,node b)
{
    double d = (a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y);
    return sqrt(d);
}

int DFS (int u)
{
    int v;
    for (v = 1;v <= m;v ++)
        if (!vis[v]&&map[u][v])
        {
            vis[v] = 1;
            if (link[v] == -1||DFS(link[v]))
            {
                link[v] = u;
                return 1;
            }
        }
    return 0;
}
int MaxMatch()
{
    int u,res = 0;
    memset (link,-1,sizeof(link));
    for (u = 1;u <= n;u ++)
    {
        memset (vis,0,sizeof(vis));
        if (DFS(u))
            res ++;
    }
    return res;
}
int main ()
{
    int s,v,dis,i,j;
    double x0,y0;
    while (~scanf ("%d%d%d%d",&n,&m,&s,&v))
    {
        memset (map,0,sizeof(map));
        dis = s*v;
        for (i = 1;i <= n;i ++)
        {
            scanf ("%lf%lf",&x0,&y0);
            g[i].x = x0;
            g[i].y = y0;
        }
        for (i = 1;i <= m;i ++)
        {
            scanf ("%lf%lf",&x0,&y0);
            h[i].x = x0;
            h[i].y = y0;
        }
        for (i = 1;i <= n;i ++)
            for (j = 1;j <= m;j ++)
                if (len(g[i],h[j]) < dis)
                    map[i][j] = 1;
        int ans = n - MaxMatch();
        printf ("%d\n",ans);
    }
    return 0;
}