poj2253Frogger 最短路水题

//一个无向图，求从起点到终点的所经过的路径中的最大值最小
//最短路，更新的时候dis[i] = min(dis[i] , max(dis[u] , map[u][i]));
#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#include<cmath>
using namespace std ;
const int maxn = 210 ;
const int inf = 0x3f3f3f3f ;
double map[maxn][maxn];
int vis[maxn] ;
double dis[maxn] ;int n ;
double x[maxn] , y[maxn] ;
void spfa()
{
    for(int i = 2;i <= n;i++)
    dis[i] = inf ;
    dis[1] = 0 ;
    queue<int> que ;
    que.push(1) ;
    memset(vis ,0  ,sizeof(vis)) ;
    while(que.size())
    {
        int u = que.front() ;
        que.pop() ;
        vis[u] = 0 ;
        for(int i = 1;i <= n;i++)
        if(dis[i] > max(dis[u] , map[u][i]))
        {
            dis[i] = max(dis[u] , map[u][i]) ;
            if(!vis[i])
            vis[i] = 1 , que.push(i) ;
        }
    }
    printf("Frog Distance = %.3lf\n" , dis[2]) ;
    puts("") ;
}
int main()
{
    //freopen("in.txt" ,"r" , stdin) ;
    int cas =  0;
    while(scanf("%d" , &n) && n)
    {
        memset(map , 0 , sizeof(map)) ;
        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\n" , ++cas) ;
        spfa() ;
    }
    return 0 ;
}