poj 2253 （方法二）dijkstra变形 路径最大边的最小值

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

2
0 0
3 4

3
17 4
19 4
18 5

0

Sample Output

Scenario #1
Frog Distance = 5.000

Scenario #2
Frog Distance = 1.414

要求：给出n个点的坐标，求出第1个点到第2个点的多个路径中的最大边的最小值。

方法：dijkstra变形

1.dijkstra中dis数组存的是第1个点与第i个点的最短距离。然后用已知的最小未遍历的dis[i]去更新其他的dis[j]，更新语句是

dis[j] = min(dis[j] , dis[i] + map1[i][j])。

2.此题变形为dis数组存的是第1个点与第i个点的多个路径中的最大边的最小值。然后用已知的最小未遍历的dis[i]去更新其他的dis[j]，更新语句是dis[j] = min(dis[j] , max(dis[i] , map1[i][j]))。
 

#include<iostream>
#include<cmath>
#include<string>
using namespace std;
const double inf=9999999;
const int N=220;
double G[N][N];
int n,id,vst[N];
struct node{
	int x,y;
	node(){
	}
	node(int xx,int yy){
		x=xx;
		y=yy;
	}
}nd[N];
double dst[N];
void dijkstra(int s){
	for(int i=1;i<=n;i++){
		dst[i]=inf;
		vst[i]=0;
	}
	dst[s]=0;
	for(int i=0;i<n;i++){
		int v;
		double min_d=inf;
		for(int j=1;j<=n;j++){
			if(!vst[j]&&dst[j]<min_d){
				v=j;
				min_d=dst[j];
			}
		}
		vst[v]=1;
		for(int i=1;i<=n;i++){
			dst[i]=min(dst[i],max(dst[v],G[v][i]));
		}
	}
	printf("Scenario #%d\n",id);
	printf("Frog Distance = %.3f\n\n",dst[2]);
}
int main(){
	id=0;
	while(~scanf("%d",&n)){
		if(n==0) break;
		id++;
		for(int i=1;i<=n;i++){
			G[i][i]=0;
		}
		for(int i=1;i<=n;i++){
			scanf("%d%d",&nd[i].x,&nd[i].y);
		}
		for(int i=1;i<=n;i++){
			for(int j=i+1;j<=n;j++){
				double xx=nd[i].x-nd[j].x;
				double yy=nd[i].y-nd[j].y;
				G[i][j]=G[j][i]=sqrt(xx*xx+yy*yy);
			}
		}
		dijkstra(1);
		
	}
	return 0;
}