POJ2253 Frogger

Frogger
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 41297 Accepted: 13245
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：两个青蛙在溪流的石头上，现在青蛙1想到青蛙2那里去，求他们之间所有路中最长边中的最短边；
首先dijkstra

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int inf=0x3f3f3f3f;
const int maxn=205;
int n;
double map[maxn][maxn],x[maxn],y[maxn];
double dis[maxn];
bool vis[maxn];
void dijkstra(int st,int ed)
{
    memset(vis,0,sizeof(vis));
    vis[st]=1;
    for(int i=1;i<=n;i++)
        dis[i]=map[i][st];
    for(int i=1;i<n;i++)
    {
        int point=0;
        double minn=inf;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j]&&dis[j]<minn)
                minn=dis[j],point=j;//这个是在各个通路上求最大值
        }
        vis[point]=1;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j]&&dis[j]>max(dis[point],map[point][j]))//这是求最大值的最小值，大于号的意思就是这样。dis[j]就不是起点到终点的最大值了，而是一条路上的最大值；
                dis[j]=max(dis[point],map[point][j]);
        }
    }
    printf("Frog Distance = %.3lf
",dis[ed]);
}
int main()
{
    int car=1;
    while(~scanf("%d",&n),n)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%lf%lf",&x[i],&y[i]);
            for(int j=1;j<i;j++)
            {
                map[i][j]=map[j][i]=sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
            }

        }
        printf("Scenario #%d
",car++);
        dijkstra(1,2);
        printf("
");
    }
    return 0;
}