POJ

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

20 03 4317 419 418 50

Sample Output

Scenario #1Frog Distance = 5.000Scenario #2Frog Distance = 1.414

   注意题意!!!!

       只要能从第一个·点跳到第二个点，无论总路程多长，都可以。然后从这些满足条件的方式中选择跳的最短的一次（注意：不是总路程最短，是需要一步跳的路程最短！！！）叙述有点绕，理解就好了。

      所以啦，不是简单的模板题，需要做些改变。看代码：

#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>#define INF  0x3f3f3f3fusing namespace std;int n;double G[360][360], dis[360];int vis[360];struct note{    double x, y;} num[360];void dijkstra(){    for(int i = 1; i <= n; i++)    {        dis[i] = G[1][i];        vis[i] = 0;    }    vis[1] = 1;    for(int i = 1; i < n; i++)    {        double min=INF;        int u=1;        for(int j = 1; j <= n; j++)        {            if(vis[j] == 0 && min > dis[j])            {                u = j;                min = dis[j];            }        }        vis[u] = 1;        for(int j = 1; j <= n; j++)        {            if(!vis[j] && G[u][j] < INF)            {                double maxn = max(dis[u],G[u][j]);                if(dis[j] > maxn)                    dis[j] = maxn;            }        }    }}int main(){    int f = 1;    while(~scanf("%d", &n)&&n)    {        for(int i = 1; i <= n; i++)            for(int j = 1; j <= n; j++)                if(i==j)G[i][j]=0;                else G[i][j]=INF;        for(int i = 1; i <= n; i++)            scanf("%lf%lf", &num[i].x, &num[i].y);        for(int i = 1; i <= n; i++)        {            for(int j = i; j <= n; j++)            {                double a = sqrt((num[i].x-num[j].x)*(num[i].x-num[j].x)+(num[i].y-num[j].y)*(num[i].y-num[j].y));                G[i][j] = a;                G[j][i] = a;            }        }        dijkstra();        printf("Scenario #%d\n",f++);        printf("Frog Distance = %.3lf\n\n",dis[2]);    }    return 0;}