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POJ2536_Gopher II(二分图最大匹配)

2024-07-14 05:53:05   221人阅读

解题报告

http://blog.csdn.net/juncoder/article/details/38156509

题目传送门

题意：

n只地鼠，m个洞，老鹰的到达地面的时间s，地鼠的移动速度v，求多少只地鼠会被老鹰吃了。

思路：

地鼠和洞看成两集合，建立二分图。只有当地鼠到洞的时间少于老鹰到地面的时间才连边。

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;
int n,m,s,v,mmap[500][500],vis[500],pre[500];
struct point
{
    double x,y;
}G[200],H[200];
double dis(point p1,point p2)
{
    return  sqrt((p2.x-p1.x)*(p2.x-p1.x)+(p2.y-p1.y)*(p2.y-p1.y));
}
int dfs(int x)
{
    for(int i=n+1;i<=n+m;i++){
        if(!vis[i]&&mmap[x][i]){
            vis[i]=1;
            if(pre[i]==-1||dfs(pre[i])){
                pre[i]=x;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    //std::ios::sync_with_stdio(false);
    int i,j,a,b,t;
    while(~scanf("%d%d%d%d",&n,&m,&s,&v)){
        memset(pre,-1,sizeof(pre));
        memset(mmap,0,sizeof(mmap));
        for(i=1;i<=n;i++){
            scanf("%lf%lf",&G[i].x,&G[i].y);
        }
        for(i=1;i<=m;i++){
            scanf("%lf%lf",&H[i].x,&H[i].y);
        }
        for(i=1;i<=n;i++){
            for(j=1;j<=m;j++){
                double d=dis(G[i],H[j]);
                if(d/v<=(double)s){
                    mmap[i][n+j]=1;
                }
            }
        }
        int ans=0;
        for(i=1;i<=n;i++){
            memset(vis,0,sizeof(vis));
            ans+=dfs(i);
        }
        printf("%d\n",n-ans);
    }
    return 0;
}

Gopher II
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 6438 Accepted: 2640

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


POJ2536_Gopher II(二分图最大匹配)


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