Frogger POJ - 2253

//给出一个无向图,求一条点1~点2的路径使得路径上的最大边权最小,注意输出G++不能用.3lf
#include<iostream>
#include<queue>
#include<cstring>
#include<cstdio>
#include<vector>
#include<cmath>
#include<iomanip>
using namespace std;
const int inf=0x3f3f3f3f;
vector<pair<int,double> >e[209];
int n,x[209],y[209];
double dis[209];
double ww(int i,int j)
{
    return sqrt(double(x[i]-x[j])*(x[i]-x[j])+double(y[i]-y[j])*(y[i]-y[j]));
}
bool vis[209];
void spfa()
{
    for(int i=1; i<=n; i++)
        dis[i]=inf;
    queue<int>q;
    q.push(1);
    dis[1]=0;
    vis[1]=1;
    while(!q.empty())
    {
        int top=q.front();
        q.pop();
        vis[top]=0;
        for(int i=0; i<e[top].size(); i++)
        {
            int temp=e[top][i].first;
            if(dis[temp]>max(dis[top],e[top][i].second))
            {
                dis[temp]=max(dis[top],e[top][i].second);
                if(!vis[temp])
                {
                    vis[temp]=1;
                    q.push(temp);
                }
            }
        }
    }
}
int main()
{
    int cntt=0;
    while(~scanf("%d",&n)&&n)
    {
        //edge ee;
        for(int i=1; i<=n; i++)
            scanf("%d%d",&x[i],&y[i]);
        for(int i=1; i<=n; i++)
            for(int j=i+1; j<=n; j++)
            {
                e[i].push_back(make_pair(j,ww(i,j)));
                e[j].push_back(make_pair(i,ww(i,j)));
            }
        spfa();
         printf("Scenario #%d\nFrog Distance = %.3lf\n\n",++cntt,dis[2]);
        //cout << "Scenario #" << ++cntt << endl;
        //cout << "Frog Distance = " <<fixed << setprecision(3) << dis[2] << endl << endl;
        for(int i=1; i<=n; i++)
         e[i].clear();
    }
    return 0;
}