[uva] 10310 Dog and Gopher

[uva] 10310Dog and Gopher

【题目】http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1251

【题意】求最短可逃跑路线

【算法】简单几何

【题解】地鼠跑的路程*2<狗狗跑的路程的最优解

【注意】这是水题

【AC代码】

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cmath>

using namespace std;

 

struct point{

    double x,y;

}a,b,p[1000];

 

double d(point p1,point p2){

    return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));

}

 

int main(){

    int n,i,t;

 

    while(scanf("%d",&n)!=EOF){

        //printf("%d\n",n);

        scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y);

        //printf("%.2f %.2f %.2f %.2f\n",a.x,a.y,b.x,b.y);

        for(i=0;i<n;i++)scanf("%lf%lf",&p[i].x,&p[i].y);

        t=-1;

 

        for(i=0;i<n;i++)if(2*d(a,p[i])<=d(b,p[i])&&(t==-1||d(a,p[t])>d(a,p[i]))) t=i;

        if(t==-1)printf("The gopher cannot escape.\n");

        else printf("The gopher can escape through the hole at (%.3f,%.3f).\n",p[t].x,p[t].y);

    }

    return 0;

}

【心得】这是开始几何的第一题么