POJ 2536 Gopher II

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Gopher II
Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 6672 Accepted: 2730

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 5 101.0 1.02.0 2.0100.0 100.020.0 20.0

Sample Output

1
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
#include <cstdio>#include <string.h>#include <stdlib.h>#include <stdio.h>#include <algorithm>#include <math.h>using namespace std;int n,m,s,v;int map[501][501],xx[501],yy[501],b[501];double xs[501];double ys[501];int dfs(int r){    for(int i=1;i<=m+n;i++)    {        if(map[r][i] && !b[i])        {            b[i]=1;            if(yy[i]==-1 || dfs(yy[i]))            {                yy[i]=r;                xx[r]=i;                return 1;            }        }    }    return 0;}int pipei(){    memset(xx,-1,sizeof(xx));    memset(yy,-1,sizeof(yy));    int ans=0;    for(int i=1;i<=n;i++)    {        if(xx[i]==-1)        {            memset(b,0,sizeof(b));            ans+=dfs(i);        }    }    return ans;}int main(){   while(~scanf("%d%d%d%d",&n,&m,&s,&v))   {       int i,j;       memset(map,0,sizeof(map));       for(i=1;i<=n;i++)        scanf("%lf%lf",&xs[i],&ys[i]);       for(i=n+1;i<=n+m;i++)        scanf("%lf%lf",&xs[i],&ys[i]);       for(i=1;i<=n;i++)       {           for(j=1+n;j<=m+n;j++)           {               if(i!=j)               {            double t=sqrt((xs[i]-xs[j])*(xs[i]-xs[j]) + (ys[i]-ys[j])*(ys[i]-ys[j]));                    if(t/v<=s)                      map[i][j]=1;               }           }       }       int ans=pipei();       printf("%d\n",n-ans);   }  return 0;}


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