Gopher II, where are the number of rats that were caught at the end ?,

Gopher II, where are the number of rats that were caught at the end ?,

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Gopher II

Sha, 10080

Time Limit: 3000 MS

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

The areNGophers andMGopher holes, each at distinct (x, y) coordinates. A hawk arrives and if a gopher does not reach a hole inSSeconds it is vulnerable to being eaten. A hole can save at most one gopher. All the gophers run at the same velocityV. The gopher family needs an escape strategy that minimizes the number of vulnerable gophers.

The input contains several cases. The first line of each case contains four positive integers less than 100:N,M,S, AndV. The nextNLines give the coordinates of the gophers; the followingMLines 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 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

Output for Sample Input

1

Question:

Give you the coordinates of the hamster and the hamster hole, the speed at which the hamster runs, and the time when the hunter arrives. How many hamsters are there at least?

Solution:

The largest Matching Algorithm in Hungary, template question.

This article is recommended: http://blog.csdn.net/dark_scope/article/details/8880547

Code:

#include<iostream>#include<cstdio>#include<cmath>using namespace std;const int maxn=110;struct node{   double x,y;   node(double x0=0.0,double y0=0.0){x=x0,y=y0;}};int n,m,k,v,pipei[maxn];bool line[maxn][maxn],visited[maxn];node dishu[maxn],holes[maxn];void input(){    double dis,t;    for(int i=0;i<=m;i++) pipei[i]=0;    for(int i=1;i<=n;i++){        scanf("%lf%lf",&dishu[i].x,&dishu[i].y);    }    for(int i=1;i<=m;i++){        scanf("%lf%lf",&holes[i].x,&holes[i].y);        for(int j=1;j<=n;j++){            dis=sqrt((holes[i].x-dishu[j].x)*(holes[i].x-dishu[j].x)+(holes[i].y-dishu[j].y)*(holes[i].y-dishu[j].y));            if(k*v<dis) line[j][i]=false;            else line[j][i]=true;        }    }}bool find(int x){    for(int j=1;j<=m;j++){        if(line[x][j]&&!visited[j]){            visited[j]=true;            if(!pipei[j]||find(pipei[j])){                    pipei[j]=x;                    return true;            }        }    }    return false;}void solve(){    int cnt=0;    for(int i=1;i<=n;i++){        for(int j=0;j<=m;j++)visited[j]=false;        if(find(i)) cnt++;    }    printf("%d\n",n-cnt);}int main(){    while(cin>>n>>m>>k>>v){        input();        solve();    }    return 0;}