POJ 2253 Frogger

标签：poj   acm   

Description

Input

Output

Sample Input

2
0 0
3 4

3
17 4
19 4
18 5

0

Sample Output

Scenario #1
Frog Distance = 5.000

Scenario #2
Frog Distance = 1.414

最短路径的变形。

AC代码如下：

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
using namespace std;

double max(double a,double b)
{
    return a>b?a:b;
}

int main()
{
    int n,b;
    int i,j,cas=0;
    double x[505],y[505],sx,sy,ex,ey,w[505][505],d[505],vis[505];
    while(cin>>n,n)
    {
        cas++;
        double ans;
        for(i=0;i<n;i++)
            {
                cin>>x[i]>>y[i];
            }
        for(i=0;i<n;i++)
            for(j=0;j<=i;j++)
        {
            w[i][j]=sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
            w[j][i]=w[i][j];
        }
        memset(vis , 0, sizeof vis);
        for(i=0;i<n;i++)
        {
            if(i==0)
                d[i]=0;
             else d[i]=0xffffffff;
        }
        for(i=0;i<n;i++)
        {
            int a;double m=0xffffffff;
            for(j=0;j<n;j++)
                if(!vis[j]&&d[j]<m)
                {m=d[j];a=j;}
            vis[a]=1;
            if(a==1)
                break;
            for(j=0;j<n;j++)
            {
                ans=sqrt((x[a]-x[j])*(x[a]-x[j])+(y[a]-y[j])*(y[a]-y[j]));
                if(!vis[j]&& d[j]>max(d[a],ans))
                {
                    d[j]=max(d[a],ans);
                }
            }
        }
        printf("Scenario #%d\nFrog Distance = %.3f\n\n",cas,d[1]);
    }
    return 0;
}

POJ 2253 Frogger,布布扣,bubuko.com

POJ 2253 Frogger

标签：poj   acm   

原文地址：http://blog.csdn.net/hanhai768/article/details/37596711