POJ2253 Frogger

Frogger

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 34865		Accepted: 11192

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. 

给定湖中两只青蛙所在的石头，以及其他石头的坐标，试计算两只青蛙之间的青蛙距离。

青蛙距离：两块石头之间所有路径中的最大跳跃距离的最小值

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.

多组数据

每组第一行为整数n，表示石头数目。接下来n行每行两个整数xi和yi，表示第i块石头坐标。

输入文件最后一行为0

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

因为已经限定了两只青蛙在石头1和2上，所以是单源最短路径问题

跑了遍floyd就过了

求青蛙距离：map[i][j]=min(map[i][j],max(map[i][k],map[k][j])

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int mxn=2000;
int x[mxn],y[mxn];//石头坐标 
double mp[210][210];//图 
int n;
double dis(int x1,int x2,int y1,int y2){//求两点间距离 
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
void cdis(){//将两点间距离存入邻接矩阵 
    memset(mp,0,sizeof(0));
    int i,j;
    for(i=1;i<=n;i++)
      for(j=1;j<=n;j++){
          mp[i][j]=dis(x[i],x[j],y[i],y[j]);
      }
    return;
}
int main(){
    int T=0;
    while(scanf("%d",&n) && n){
        printf("Scenario #%d
",++T);
        int i,j;
        for(i=1;i<=n;i++){
            scanf("%d%d",&x[i],&y[i]);
        }
        cdis();
        for(int k=1;k<=n;k++)
            for(i=1;i<=n;i++)
                  for(j=1;j<=n;j++){
                    mp[i][j]=min(mp[i][j],max(mp[i][k],mp[k][j]));
//                    printf("%d  %d  %d
",k,i,j);
//                    printf("%.3f  %.3f  %.3f
",mp[i][j],mp[i][k],mp[k][j]);
              }
        printf("Frog Distance = %.3f

",mp[1][2]);//注意空行 
    }
    return 0;
}