poj2253(dijistra变形+堆优化）

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.

 

 

这个题是找最短路 但是最短路的定义是

对于一条从起点到终点的路径 该路径的长度为该路径中最长边的长度

我们要找到最短的路径

//
// Created by xingchaoyue on 2019/5/5.
//

#include<iostream>
#include<cstdio>
#include<queue>
#include<cmath>
using namespace std;
const int maxn = 2000;
const int inf = 0x3f3f3f3f;

double mp[maxn][maxn];
double dist[maxn];
int vis[maxn];
struct node{
    double x,y;
};
node p[maxn];
int n;
struct node1{
    int v;
    double d;
     node1(int _v,double _d):v(_v),d(_d){}
     bool operator <(const node1 b)const{
         return d>b.d;
     }
};

double getdis(node a,node b){
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
void init(){
    for(int i =1;i<=n;++i){
        for(int j =1;j<=n;++j){
            mp[i][j] = inf;
        }
    }
}
void dijistra(int st,int n){
    for(int i =1;i<=n;++i){
        dist[i] = inf;
        vis[i] = 0;
    }
    dist[st] = 0;
    node1 temp(st,0);
    priority_queue<node1>q;
    q.push(temp);
    while(!q.empty()){
        node1 t = q.top();
        q.pop();
        int v = t.v;
        //cout<<v<<endl;
        double dis = t.d;
        if(vis[v])continue;
        vis[v] = 1;
        for(int i =1;i<=n;++i){
            //cout<<dist[i]<<" ";
            if(dist[i]>max(dist[v],mp[v][i]))
            {
                dist[i] = max(dist[v],mp[v][i]);
                node1 temp(i,dist[i]);
                q.push(temp);
            }
        }
        //cout<<endl;
    }

}
int main(){
    int t = 1;
    while(~scanf("%d",&n)&&n){
        init();
        for(int i =1;i<=n;++i){
            cin>>p[i].x>>p[i].y;
        }
        for(int i =1;i<=n;++i){
            for(int j =i+1;j<=n;++j){
                mp[i][j] = mp[j][i] = getdis(p[i],p[j]);
            }
        }
       /* for(int i =1;i<=n;++i){
            for(int j =1;j<=n;++j){
                cout<<mp[i][j]<<" ";
            }
            cout<<endl;
        }*/
        int st = 1;
        int ed = 2;
        dijistra(1,n);
        printf("Scenario #");
        cout<<t++<<endl;
        printf("Frog Distance = %.3f\n",dist[ed]);
        cout<<endl;

    }
}