学步园

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 ++)