poj 2536 GopherII（二分图匹配）

Description

The gopher family, having averted the canine threat, must face a new predator.

The are n gophers and m gopher holes, each at distinct (x, y)
coordinates. A hawk arrives and if a gopher does not reach a hole in s
seconds it is vulnerable to being eaten. A hole can save at most one
gopher. All the gophers run at the same velocity v. The gopher family
needs an escape strategy that minimizes the number of vulnerable
gophers.

Input

The
input contains several cases. The first line of each case contains four
positive integers less than 100: n, m, s, and v. The next n lines give
the coordinates of the gophers; the following m lines give the
coordinates of the gopher holes. All distances are in metres; all times
are in seconds; all velocities are in metres per second.

Output

Output consists of a single line for each case, giving the number of vulnerable gophers.

Sample Input

2 2 5 10
1.0 1.0
2.0 2.0
100.0 100.0
20.0 20.0

Sample Output

1

题目大意，有一块地上面有N只老鼠和M个只能装的下一只老鼠的老鼠洞，老鼠和老鼠洞的坐标已给出，老鹰在S秒后来到，老鼠的速度全部为V，问最少有多少老鼠来不及进洞
简单的二分图匹配算法，直接贴出代码：

#include<iostream>
#include<string.h>
#include<cmath>
using namespace std;
int map[][],v[];
double x[],y[],xx,yy;
int match[];
int N,M,S,V;
double disIsOK(double x1,double y1,double x2,double y2){
    return S*V>=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
int dfs(int ii){
    for(int i=;i<=N;i++){
        if(map[i][ii]&&!v[i]){
            v[i]=;
            if(match[i]==||dfs(match[i])){
                match[i]=ii;return ;
            }
        }
    }
    return ;
}
int main(){
    while(cin>>N>>M>>S>>V){
        memset(map,,sizeof(map));
        memset(match,,sizeof(match));
        for(int i=;i<=N;i++){
            cin>>x[i]>>y[i];
        }
        for(int i=;i<=M;i++){
            cin>>xx>>yy;
            for(int j=;j<=N;j++){
                if(disIsOK(x[j],y[j],xx,yy)){
                    map[j][i]=;
                }
            }
        }
        int res=;
        for(int i=;i<=M;i++){
            memset(v,,sizeof(v));
            if(dfs(i))res++;
        }
        cout<<N-res<<endl;
    }
    return ;
}

结果截图：