F - Frogger

题目：

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：

一只青蛙想要跳到另一只青蛙那里，让你求出来最短路中的最大的跳跃距离；

给你一个n，代表有n个石头的坐标，接下来有n 对x，y；

其中第一对和第二对数据是两个青蛙的位置，剩下的是别的石头的位置，让你求出来结果；

思路：

在最短路中顺便判断青蛙最大的跳跃距离；

注意：

坐标的位置是两个整数，不能用double，但是可以用int和float，不晓得为什么，跪求大佬指点。。。。。

代码如下：

#include
#include
#define N 1010
#include
#include
using namespace std;

int n;
double e[N][N];

struct node//存储坐标；
{
    int a,b;
} s[N];

int main()
{
    int kk=1;
    while(~scanf("%d",&n)&&n)
    {
        int i,j,k;
        for(i=1; i<=n; i++)
        {
            scanf("%d%d",&s[i].a,&s[i].b);
        }
        memset(e,0x3f,sizeof e);
        for(i=1; i<=n; i++)
        {
            for(j=1; j<=n; j++)
            {//计算距离；
                e[i][j]=sqrt(1.0*(s[i].a-s[j].a)*(s[i].a-s[j].a)+(s[i].b-s[j].b)*(s[i].b-s[j].b));
            }
        }
        //Floyd-Warshall
        for(k=1; k<=n; k++)
        {
            for(i=1; i<=n; i++)
            {
                for(j=1; j<=n; j++)
                {
                    e[i][j]=min(e[i][j],max(e[i][k],e[k][j]));
                }
            }
        }
        printf("Scenario #%d\n",kk++);
        printf("Frog Distance = %.3lf\n\n",e[1][2]);//注意输出格式；
    }
    return 0;
}