Frogger

Description

Freddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on another stone. He plans to visit her, but since the water is dirty and full of tourists' sunscreen, he wants to avoid swimming and instead reach her by jumping. 
Unfortunately Fiona's stone is out of his jump range. Therefore Freddy considers to use other stones as intermediate stops and reach her by a sequence of several small jumps. 
To execute a given sequence of jumps, a frog's jump range obviously must be at least as long as the longest jump occuring in the sequence. 
The frog distance (humans also call it minimax distance) between two stones therefore is defined as the minimum necessary jump range over all possible paths between the two stones. 

You are given the coordinates of Freddy's stone, Fiona's stone and all other stones in the lake. Your job is to compute the frog distance between Freddy's and Fiona's stone. 

Input

The input will contain one or more test cases. The first line of each test case will contain the number of stones n (2<=n<=200). The next n lines each contain two integers xi,yi (0 <= xi,yi <= 1000) representing the coordinates of stone #i. Stone #1 is Freddy's stone, stone #2 is Fiona's stone, the other n-2 stones are unoccupied. There's a blank line following each test case. Input is terminated by a value of zero (0) for n.

Output

For each test case, print a line saying "Scenario #x" and a line saying "Frog Distance = y" where x is replaced by the test case number (they are numbered from 1) and y is replaced by the appropriate real number, printed to three decimals. Put a blank line after each test case, even after the last one.

Sample Input

2

0 0

3 4

3

17 4

19 4

18 5

Sample Output

Scenario #1

Frog Distance = 5.000

Scenario #2

Frog Distance = 1.414

Source

Ulm Local 1997

题目大意：定义“青蛙距离”为Freddy跳到Fiona那里所需要的若干次跳跃中最长的那一次。第一行有一个整数n(2<=n<=200)，表示湖中一共有多少块石头。接下来的n行，每一行有两个整数xi，yi(0 <= xi,yi <= 1000)，表示第i块石头的坐标。第1块石头的坐标是Freddy所在的位置，第二块石头的坐标是Fiona所在的位置，其他的石头上都没有青蛙。求青蛙距离。

#include<stdio.h>
#include<limits.h>
#include<math.h>
double firstjump[205];
double map[205][205];
int check[205];											//用于标记第i块石头是否被访问 
int x[205],y[205],n;
double max,min;
void sove()
{
	firstjump[0]=0;
	check[0]=1;
	int begin=0;
	while(begin!=1)
	{
        min=INT_MAX;
        for(int i=0;i<n;i++)
        if(!check[i]&&firstjump[i]<min)
        {
             min=firstjump[i];
             begin=i;										//记录第一次跳跃后青蛙所在位置 
        }
        if(min>max)  
		max=min;
        for(int i=0;i<n;i++)
        {
            if(!check[i]&&map[begin][i]<INT_MAX&&firstjump[i]>map[begin][i])
            firstjump[i]=map[begin][i];
            check[begin]=1;
        }	
	}
}
int main()
{
	int cas=0;
	while(scanf("%d",&n),n!=0)
	{
		cas++;
		max=INT_MIN;
		for(int i=0;i<n;i++)
		scanf("%d %d",&x[i],&y[i]);
		for(int i=0;i<n;i++)						// 初始化第i块跳到第j块的距离 
		for(int j=0;j<n;j++)
		{
			map[i][j]=INT_MAX;
		}
        for(int i=0;i<n;i++)						//计算第i块跳到第j块的距离 
        for(int j=0;j<n;j++)
        {
           map[i][j]=sqrt((1.0*x[i]-1.0*x[j])*(1.0*x[i]-1.0*x[j])+(1.0*y[i]-1.0*y[j])*(1.0*y[i]-1.0*y[j]));		
        }
        for(int i=0;i<n;i++)						//第一次跳跃的最短距离 
        {
        	firstjump[i]=map[0][i];
        	check[i]=0;
        }
        sove();
        printf("Scenario #%d\n",cas);
        printf("Frog Distance = %.3f\n",max);
        printf("\n");
	}
	return 0;
}