POJ 2253 Frogger (求每条路径中最大值的最小值，Dijkstra变形)

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<math.h>
using namespace std;

const int MAXN = 210;

pair<int,int>p[MAXN];

double getdist(pair<int,int>p1,pair<int,int>p2)
{
    return sqrt((double)(p1.first-p2.first)*(p1.first-p2.first)+(p1.second-p2.second)*(p1.second-p2.second));
}

const double INF = 1e20;
double cost[MAXN][MAXN];
double dis[MAXN];
bool vis[MAXN];

void Dijkstra(int n,int st)
{
    for(int i=1;i<=n;i++){
        dis[i]=INF;
        vis[i]=false;
    }
    dis[st]=0;
    for(int j=0;j<n;j++){
        int k=-1;
        double Min=INF;
        for(int i=1;i<=n;i++){
            if(!vis[i]&&Min>dis[i]){
                Min=dis[i];
                k=i;
            }
        }
        if(k==-1) break;
        vis[k]=true;
        for(int i=1;i<=n;i++)
            if(!vis[i]&&max(dis[k],cost[k][i])<dis[i])
                dis[i]=max(dis[k],cost[k][i]);
    }
}

int main()
{
    int n;
    int x,y;
    int iCase=0;
    while(scanf("%d",&n)==1&&n){
        iCase++;
        for(int i=1;i<=n;i++){
            scanf("%d%d",&x,&y);
            p[i]=make_pair(x,y);
        }
        for(int i=1;i<=n;i++){
            for(int j=i;j<=n;j++){
                if(i==j) cost[i][j]=0.0;
                else cost[i][j]=cost[j][i]=getdist(p[i],p[j]);
            }
        }
        Dijkstra(n,1);
        printf("Scenario #%d\n",iCase);
        printf("Frog Distance = %.3f\n\n",dis[2]);

    }
    return 0;
}