POJ 2253 Frogger

	Frogger

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

Source

Ulm Local 1997


/*
       题目大意是给予N块石头，一只名字叫Freddy的青蛙坐在编号为1的石头上，他想去找坐在编号2石头上的Fiona去玩。但是水太脏了，所以他想跳过去。问在1到N的石头上青蛙Freddy所要跳的最短距离是多少，要找尽量近的石头去跳。

      石头数量小于200，直接用floyd算法求最短路的最短边就可以。第二种方法是以编号为1的石头建最小生成树，遍历得到生成树中最小的边权值也可。

*/

#include <algorithm>
#include <queue>
#include <cstring>
#include <cstdio>
#include <vector>
#include <stack>
#include <string>
#include <cmath>
const int INF = 0x3f3f3f3f;
using namespace std;

double Map[220][220];
double A[220][220];
struct node
{
    double x,y;
}a[220];

double floyd(int n)
{
    for(int i = 1;i<=n;i++)
        for(int j=1;j<=n;j++)
            A[i][j] = Map[i][j];
    for(int k=1;k<=n;k++)
    {
        for(int i=1;i<=n-1;i++)
        {
            for(int j=i+1;j<=n;j++)
            {
                if(A[i][k] < A[i][j] && A[k][j] < A[i][j])
                {
                    if(A[i][k] < A[k][j])
                        A[i][j] = A[j][i] = A[k][j];
                    else
                        A[i][j] = A[j][i] =A[i][k];
                }
            }
        }
    }
    return A[1][2];
}
double f(double x1,double y1,double x2,double y2 )
{
    return (x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);
}

int main()
{
    int n,t = 1;
    while(~scanf("%d",&n) && n)
    {
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                i==j?Map[i][j]=0.0:Map[i][j] = (double)INF;
        for(int i=1;i<=n;i++)
        {
            scanf("%lf %lf",&a[i].x,&a[i].y);
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=i+1;j<=n;j++)
            {
                Map[i][j] = Map[j][i] = sqrt(f(a[i].x,a[i].y,a[j].x,a[j].y));
            }
        }
        printf("Scenario #%d\nFrog Distance = %.3f\n\n",t++,floyd(n));
    }
    return 0;
}