Gopher II（POJ-2536 ）

Problem 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

题意：有 n 个地鼠和 m 个洞，每个洞只能容纳 1 只地鼠，当鹰飞来时，n 个地鼠若能在 s 秒内从当前位置回到洞中，就能不被鹰抓住，每只地鼠的速度为 v，求被抓住的地鼠个数的最小值

思路：将地鼠与洞看做二分图，如果一个地鼠能在 s 秒内到达洞，那么就在其之间建立一条边，根据题意，被抓住的个数=n-最大匹配数

Source Program

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define PI acos(-1.0)
#define E 1e-6
#define MOD 16007
#define INF 0x3f3f3f3f
#define N 10001
#define LL long long
using namespace std;
bool vis[N];
int link[N];
vector<int> G[N];
bool dfs(int x){
    for(int i=0;i<G[x].size();i++){
        int y=G[x][i];
        if(!vis[y]){
            vis[y]=true;
            if(link[y]==-1 || dfs(link[y])){
                link[y]=x;
                return true;
            }
        }
    }
    return false;
}
int hungarian(int n)
{
    int ans=0;
    for(int i=1;i<=n;i++){
        memset(vis,false,sizeof(vis));
        if(dfs(i))
            ans++;
    }
    return ans;
}
struct Point{
    double x;
    double y;
}P1[N],P2[N];
double get_dis(int a,int b){
     return sqrt( (P1[a].x-P2[b].x)*(P1[a].x-P2[b].x)+(P1[a].y-P2[b].y)*(P1[a].y-P2[b].y) );
}
int main(){
    int n,m,s,v;
    while(scanf("%d%d%d%d",&n,&m,&s,&v)!=EOF&&n){
        memset(link,-1,sizeof(link));
        for(int i=0;i<N;i++)
            G[i].clear();

        for(int i=1;i<=n;i++)
            scanf("%lf%lf",&P1[i].x,&P1[i].y);
        for(int i=1;i<=m;i++)
            scanf("%lf%lf",&P2[i].x,&P2[i].y);

        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                if(get_dis(i,j)<=s*v){
                    G[i].push_back(j);
                }
            }
        }
        printf("%d\n",n-hungarian(m));

    }
    return 0;
}