http://acm.hdu.edu.cn/webcontest/contest_showproblem.php?cid=963&pid=1019&ojid=1

这一题是dijstra的变种，，，用的是dijstra的思想和方法，让求的是从一点到另一点的最大的最小值，，，，一开始木有认真读题，，，贡献了5次wa，，，

#include <iostream>
#include <cmath>
#include<cstdio>
using namespace std;
#define MAX 201
#define INF 10000.0f
float matrix[MAX][MAX];
int s[MAX];
float dis[MAX];
typedef struct point
{
    int x;
    int y;
}point;
void Dijkstra(int &n)
{         int now=1;
    for(int i = 1; i <= n; ++i)
    {
        dis[i] = matrix[now][i];
        s[i] = 0;
     }
    s[now] = 1;
    for(int i = 1; i <= n-1; ++i)
    {
        float minDis = INF;
        for(int j = 1; j <= n; ++j)
            if(!s[j] && dis[j] < minDis)
              minDis = dis[now=j];
            s[now] = 1;
         for(int j = 1; j <= n; ++j)
            if(!s[j] && matrix[now][j]!= INF)
            dis[j] = min(dis[j],max(dis[now],matrix[now][j]));
    }
}
int main()
{
    int n;
    point p[MAX];
    int count = 0;
    while(~scanf("%d",&n)&& n)
    {for(int i = 1; i <= n; ++i)
        ~scanf("%d%d", &p[i].x, &p[i].y);
        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= n; ++j)
            if(i!=j)  matrix[i][j] = INF;
              else    matrix[i][j] = 0;
            for(int i = 1; i < n; ++i)
            for(int j = i+1; j <=n; ++j)
    matrix[j][i] = matrix[i][j] = sqrt(((float)(p[i].x-p[j].x)*(p[i].x-p[j].x))
                    + (p[i].y-p[j].y)*(p[i].y-p[j].y));
                    
        Dijkstra(n);
        printf("Scenario #%d\nFrog Distance = %.3f\n\n", ++count, dis[2]);
    }
    return 0;
}