Frogger （迪杰斯拉变形）

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

0

Sample Output

Scenario #1
Frog Distance = 5.000

Scenario #2
Frog Distance = 1.414

题目大意： 输入的第一行代表石头的个数，当个数为0时结束，接着有n行，其中第2，3行分别代表A，B青蛙的坐标，其他n-2行分别代表空的石头的坐标，输出一个小数（保留三位），注意每输出一个答案还要再空一行。给出青蛙A,B和若干石头的坐标，现青蛙A想到青蛙B的位置，A可通过任意石头到达B,问从A到B多条路径中的最长边中的最短距离。

代码；

#include 
#include 
#include 
#define inf 99999999
int vis[1010];
int x[1010],y[1010];
double dis[1010];
double map[1010][1010];

double max0(double a,double b)
{
	if(a>b)
		return a;
	return b;
}
int main()
{
	int i,j,count=1,min,n,u,v;
	while(scanf("%d",&n)!=EOF)
	{
		if(n==0)
			break;
		memset(map,0,sizeof(0));
		memset(vis,0,sizeof(vis));	
		for(i=1;i<=n;i++)
			scanf("%d%d",&x[i],&y[i]);
		for(i=1;i<=n;i++)
			for(j=i+1;j<=n;j++)
				map[i][j]=map[j][i]=(double)sqrt((fabs(x[i]-x[j])*fabs(x[i]-x[j])+fabs(y[i]-y[j])*fabs(y[i]-y[j]))*1.0);
				
		for(i=1;i<=n;i++)
			dis[i]=map[1][i];
		vis[1]=1;
		for(i=1;i<=n;i++)
		{
			min=inf;
			for(j=1;j<=n;j++)
				if(vis[j]==0&&dis[j]max0(dis[u],map[u][v]))
					dis[v]=max0(dis[u],map[u][v]);  
		}
		printf("Scenario #%d\n",count++);
		printf("Frog Distance = %.3f\n\n",dis[2]);
	}
	return 0;
}