Poj_2536 Gopher II -二分图建图

题目：有m个洞，n知动物，每个洞容一知动物，问当天敌来后最少有多少知动物不能逃亡。

没看清楚最后的输出，被坑了几发

/************************************************
Author        :DarkTong
Created Time  :2016/7/31 16:12:15
File Name     :Pos_2536.cpp
*************************************************/

//#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
#define min(a, b) a<b?a:b
#define max(a, b) a>b?a:b
#define esp 1e-9
const int maxn = 100 + 10;
int w[maxn][maxn], n, m;
int Left[maxn];
bool used[maxn];
bool match(int j)
{
    for(int i=1;i<=n;++i) if(w[i][j]&&!used[i])
    {
        used[i] = true;
        if(!Left[i]||match(Left[i]))
        {
            Left[i] = j;
            return true;
        }
    }
    return false;
}
//返回最大匹配数
int hungary()
{
    int res=0;
    memset(Left, 0, sizeof(Left));
    for(int i=1;i<=m;++i)
    {
        memset(used, 0, sizeof(used));
        if(match(i)) res++;
    }
    return res;
}
double gx[maxn], gy[maxn], hx[maxn], hy[maxn];

int main()
{
    int T, cas=1;
    int s, v;
    while(scanf("%d%d%d%d", &n, &m, &s, &v)==4)
    {
        memset(w, 0, sizeof(w));

        for(int i=1;i<=n;++i) scanf("%lf%lf", &gx[i], &gy[i]);
        for(int i=1;i<=m;++i) scanf("%lf%lf", &hx[i], &hy[i]);
        for(int i=1;i<=n;++i)
            for(int j=1;j<=m;++j)
            {
                double l = (hx[j]-gx[i])*(hx[j]-gx[i]) +(hy[j]-gy[i])*(hy[j]-gy[i]);
                l = sqrt(l);
                if((double)s*v>=l) w[i][j]=1;
            }
        printf("%d\n", n-hungary());
    }

    
    return 0;
}