Gopher II UVA 10080，最后被抓的地鼠有多少只？

标签：uva

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Gopher II

UVA, 10080

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.

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 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

Output for Sample Input

1

题目大意：

给你地鼠和地鼠洞的坐标，还有地鼠奔跑的速度，以及猎人到达的时间，求最后被抓走的地鼠最少有多少只？

解题思路：

匈牙利算法最大匹配，模板题。

推荐这片文章：http://blog.csdn.net/dark_scope/article/details/8880547

代码：

#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;
}

Gopher II UVA 10080，最后被抓的地鼠有多少只？

标签：uva

原文地址：http://blog.csdn.net/hush_lei/article/details/41551655