POJ [2253] Frogger 【The shortest path deformation】

Description

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.

Sample Input

Sample Output

Scenario #1
Frog Distance = 5.000

Scenario #2
Frog Distance = 1.414

The main idea:

Given the coordinates of n points, form an undirected graph, find a path of 1 ~ 2, so that the maximum edge weight on the path is the minimum value of the maximum edge weight in all paths of 1 ~ 2.

analysis:

This question is the deformation of the shortest path, modify the definition of the dis array in Dijkstra's algorithm, save the minimum value of the maximum edge weight in all paths from the starting point to the current point in the dis array, and use the dis of the current point and the maximum of the adjacent edges Value to update the dis value of the adjacent point.

See the code for specific explanation.

//#include <bits/stdc++.h>
#include <iostream>
#include <cstring>
#include <queue>
#include <map>
#include <cmath>
#include <vector>
#include <iomanip>

using namespace std;

const int maxn=205;
const int INF=0x3f3f3f3f;

int vis[maxn];
double dis[maxn],tm[maxn];

int head[maxn];
int ans;
int e,n;

struct node{
 
    int u,v;
    
    double w;
 
    int next;
 
}edge[maxn*maxn];

void add(int u,int v,double w){//向邻接表中加边 
 
    edge[ans].u=u;
 
    edge[ans].v=v;
 
    edge[ans].w=w;
 
    edge[ans].next=head[u];
 
    head[u]=ans++;
 
}

struct dot{	//存储点的坐标 
	int x,y;
	dot(){}
	dot(int x,int y):x(x),y(y) {}
}; 

dot po[205];

struct point{	//定义优先队列中的节点 
    int id;
    double val;
    point(int id,double val):id(id),val(val) {}
    bool operator <(const point &x)const{
        return val>x.val;
    }
};

void dijkstra(int s){
    memset(vis,0,sizeof(vis));
    for(int i=1; i<=n; i++){
    	dis[i]=INF;
	}
    priority_queue<point> q;
    q.push(point(s,0));
    vis[s]=1;
    dis[s]=0;
    while(!q.empty()){
        int cur=q.top().id;
        q.pop();
        vis[cur]=1;	//优先队列中最小的值不可能再被经过其他节点的路径更新 
        if(cur==2) return;	//dis[2]算完即可退出； 
        for(int i=head[cur]; i!=-1; i=edge[i].next){
            int id=edge[i].v;
            if(!vis[id] && max(dis[cur],edge[i].w) < dis[id]){	//用当前点的dis和相邻边的最大值去更新相邻点的dis值
                dis[id]=max(dis[cur],edge[i].w);
                q.push(point(id,dis[id]));
            }
        }
    }
}

int main(){
	int cas=0;
	while(scanf("%d",&n)!=EOF&&n){
		cas++;
		ans=0;
		memset(head,-1,sizeof(head));
		int tx,ty;
		for(int i=1;i<=n;i++){
			scanf("%d%d",&tx,&ty);
			po[i]=dot(tx,ty);
		}
		//建图 
		for(int i=1;i<=n;i++){
			for(int j=i+1;j<=n;j++){
				double dist=sqrt((double)(po[i].x-po[j].x)*(po[i].x-po[j].x)+(po[i].y-po[j].y)*(po[i].y-po[j].y));
				add(i,j,dist);
				add(j,i,dist);
			}
		}
		dijkstra(1);
		printf("Scenario #%d\nFrog Distance = %.3f\n\n", cas, dis[2]);	//注意要多输出一个空行 
	}
	return 0;
}

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