Frogger(最短路_floyd变形)

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

0

Sample Output
Scenario #1
Frog Distance = 5.000

Scenario #2
Frog Distance = 1.414

题意：其实就是石头0上的青蛙陶兄看上了石头1上的青蛙尹小姐，然后陶兄想去找尹小姐，但是奈何湖里垃圾太多过不去咋办，聪明的陶兄就想了个很好的办法找别的石头为中转站过去，求陶兄跳到尹小姐的过程中最短路径中最大步伐中的最小长度。
ps：英语渣比用了很长时间才搞懂到底是啥意思。很内伤。可能你看到这还不明白，请看下图，本屌字写的很烂，凑合看勿喷。



#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
struct node
{
    double x,y;
}edge[1010];
double map[1010][1010];
double distan(struct node a,struct node b)
{
     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int n,i,j,k;
void floyd()
{
    for(k=0;k<n;k++)
        for(i=0;i<n;i++)
        for(j=0;j<n;j++)
    {
        map[i][j]=min(map[i][j],max(map[i][k],map[k][j]));//最关键的地方，也是变形的地方
    }

}
int main()
{
    int cnt=1;
    while(~scanf("%d",&n))
    {
        if(n==0)
            break;
        for(i=0;i<n;i++)
            scanf("%lf %lf",&edge[i].x,&edge[i].y);
        for(i=0;i<n;i++)
            for(j=0;j<n;j++)
            map[i][j]=distan(edge[i],edge[j]);
        floyd();
        printf("Scenario #%d\n",cnt++);
        printf("Frog Distance = %.3lf\n\n",map[0][1]);
    }
    return 0;
}