poj 2253 Frogger

Frogger

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 37217 Accepted: 11974

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

20 03 4317 419 418 50

Sample Output

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

/*

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。

*/

这一题的意思是让你找出所有路径中最大边的最小值

也就是说比如

1到2有三条路径

长度分别是

5 1 4

4 3 4

8 4 1

三条路的最大边的长度分别为5 4 8，找出最大边的最小值那就是4了，答案就是4了

#include <stdio.h>

#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#define MAX 0x3f3f3f3f
using namespace std;
double x[1002],y[1002];
double a[1002][1002];
int main()
{
int cnt = 1;
int n,j,i;
    while(~scanf("%d",&n)&&n != 0)
    {
        for(i = 0;i < n; i++)
        {
            for(j = 0;j < n; j++)
            {
                if(i == j)
                    a[i][j] = 0;
                else
                    a[i][j] = MAX;
            }
        }


        for(i = 0;i < n; i++)
        {
            scanf("%lf %lf",&x[i],&y[i]);
        }
        for(i = 0;i < n; i++)
        {
            for(j = 0;j < n; j++)
            {
                a[i][j] = sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));    //点与点的边长
            }
        }
        int k;
        for(k=0; k<n; k++)
            for(i=0; i<n; i++)
                for(j=0; j<n; j++)
                {
                    a[i][j]=min(a[i][j],max(a[i][k],a[k][j]));       //最大长度为以前的a[i][j],因为现在要更新所以要拿以前的与刚刚更新的两条路径作比较。
                }
        printf("Scenario #%d\n",cnt++);
        printf("Frog Distance = %.3f\n\n",a[0][1]);    //这里千万要用%.3f,不要用%.3lf不知道为什么用后者就是错，，，，（不明真相的吃瓜群众）
    }

}

代码菜鸟，如有错误，请多包涵！！！