poj 2253 Frogger

                                                                                     Frogger

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 55250		Accepted: 17412

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

#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cstring>
#include<vector>
#include<cmath>
#define maxn 205
using namespace std;
double dis[maxn][maxn];
int n;  //表示石头的个数,即坐标点的个数;
pair<int, int> point[maxn]; //用来存输入的各个点的坐标;
typedef pair<int, int> P;

double distan(P a, P b){  //求两坐标点之间的距离;
    double ans;
    ans = sqrt((double)(a.first-b.first)*(a.first-b.first)+(a.second-b.second)*(a.second-b.second));
    return ans;
}

void floyd_exchange(){   //改变了floyd的判断条件,不再是dis[i][i]=min(dis[i][j],dis[i][k]+dis[k][j]);
    for(int k = 0; k < n; k++){    //这个循环的顺序要注意,k在最外层;
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++){
                if(dis[i][j] > max(dis[i][k], dis[k][j]))  //而是如果从i到j可以直达,即dis[i][j],同时还存在i可以借助k点到j的路径
                    dis[i][j] = max(dis[i][k], dis[k][j]);  //那么我们根据题意找到所有能从i到j的路径中，每条路径都要跨越的两个石
            }                                               //头间最大值，在这每条路径的最大值中我们找出那条路径的最大值最小,dis[i][j]就存这个值;
        }
    }
    return;
}

int main(){
    int x, y;
    int cas = 1;
    while(~scanf("%d", &n) && n){
        printf("Scenario #%d\n", cas++);
        for(int i = 0; i < n; i++){
            scanf("%d%d", &x, &y);  //输入点的坐标;
            point[i] = make_pair(x, y);  //存储点的坐标;
        }
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++){  //计算并存储任意两点之间的距离;
                if(i == j){  //如果是同一个点
                    dis[i][j] = 0.0;
                }
                else{  //对于不同的点,由于是青蛙跳石头不用管方向,故是无向图;
                    dis[i][j] = dis[j][i] = distan(point[i], point[j]);
                }
            }
        }
        floyd_exchange();  //floyd一下;
        printf("Frog Distance = %.3lf\n\n", dis[0][1]);  //0点和1点之间的dis值;
    }
    return 0;
}