2536 Gopher II

Gopher II

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 3494 Accepted: 1461

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

Source

Waterloo local 2001.01.27

 

 

 

 

 

 

#include<stdio.h>

#include<string.h>

int usedif[100];

//usedif[i]记录Y顶点子集中编号为i的顶点是否使用,注意Y子集中的最多顶点数为(7-1)*12+12=84

int link[100];//link[i]记录与Y顶点子集中编号为i的顶点相连的X顶点子集中x的编号

int mat[305][100];//mat[i][j]表示顶点i与j之间是否有边

int gx,gy;//gx为X顶点子集中的顶点数目,gy为Y顶点子集中的顶点数目

double go[100][2],ve[100][2];

bool can(int t) //判断X中的顶点t在Y中是否有顶点与之匹配

{

    for(int i=0; i<gy; i++)

    {

        if(usedif[i]==0&&mat[t][i]) //Y中的顶点i未匹配且t与i之间有边

        {

            usedif[i]=1;

            if(link[i]==-1||can(link[i])) //link[i]=-1表示顶点i还未匹配

            {

                link[i]=t;

                return true;

            }

        }

    }

    return false;

}

int MaxMatch()

{

    int num=0;

    memset(link,-1,sizeof(link));

    for(int i=0; i<gx; i++) //对X中的每个顶点在Y中寻找与其匹配的顶点

    {

        memset(usedif,0,sizeof(usedif));

        if(can(i))  num++;

    }

    return num;//返回最大匹配数

}

int main()

{

    int n,m,s,v;

    while(scanf("%d%d%d%d",&n,&m,&s,&v)!=EOF)

    {

        gx=n;

        gy=m;

        memset(mat,0,sizeof(mat));

        for(int i=0;i<n;i++)  scanf("%lf%lf",&go[i][0],&go[i][1]);

        for(int i=0;i<m;i++) scanf("%lf%lf",&ve[i][0],&ve[i][1]);

        int d=s*v*s*v;

        for(int i=0;i<n;i++)

          for(int j=0;j<m;j++)

             if((go[i][0]-ve[j][0])*(go[i][0]-ve[j][0])+(go[i][1]-ve[j][1])*(go[i][1]-ve[j][1])<=d)  mat[i][j]=1;

        printf("%d/n",n-MaxMatch());

    }

    return 0;

}