8.17 H – Frogger

H– Frogger

 

Freddy Frog is sitting on a stone in the middle of alake. Suddenly he notices Fiona Frog who is sitting on another stone. He plansto visit her, but since the water is dirty and full of tourists' sunscreen, hewants to avoid swimming and instead reach her by jumping. 
Unfortunately Fiona's stone is out of his jump range. Therefore Freddyconsiders to use other stones as intermediate stops and reach her by a sequenceof several small jumps. 
To execute a given sequence of jumps, a frog's jump range obviously must be atleast as long as the longest jump occuring in the sequence. 
The frog distance (humans also call it minimax distance) between two stonestherefore is defined as the minimum necessary jump range over all possiblepaths between the two stones. 

You are given the coordinates of Freddy's stone, Fiona's stone and all otherstones in the lake. Your job is to compute the frog distance between Freddy'sand Fiona's stone. 

Input

The input will contain one or more test cases. The firstline 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 #2is Fiona's stone, the other n-2 stones are unoccupied. There's a blank linefollowing 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 bythe test case number (they are numbered from 1) and y is replaced by theappropriate real number, printed to three decimals. Put a blank line after eachtest 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

 

题意：一直青蛙在石头上跳，输入一组石头的位置坐标，从第一个石头跳到第二个石头上，但如果两个石头相距太远的话青蛙不想跳，找到一条路径能让青蛙每次跳动距离最小，输出这条路径中最大的距离。

#include<stdio.h>#include<algorithm>using namespace std;#include<string.h>#include<math.h>struct edge {int u,v;double w;}e[1200];bool cmp(edge a,edge b){return a.w<b.w;}int m,n,l,z=0;int p[1200];double a[1200][1200],b[1200],d[1200],s[1200][5],max=1e9;void Uset(){for(int i=1;i<=n;i++){p[i]=-1;}}int find(int x){int s;for(s=x;p[s]>=0;s=p[s]);while(s!=x){int t=p[x];p[x]=s;x=t;}return s;}void Union(int u ,int v){int t1=find(u);int t2=find(v);if(t1==t2)return ;int tp=p[t1]+p[t2];if(p[t1]>p[t2]){p[t1]=t2;p[t2]=tp;}else{p[t2]=t1;p[t1]=tp;}}double min(double a,double b){if(a<b)a=b;return a;}void Kruskal(){double sum=-1;Uset();for(int i=0;i<l;i++){if(find(e[i].u)!=find(e[i].v)){Union(e[i].u,e[i].v);sum=min(e[i].w,sum);if(find(1)==find(2))break;}}z++;printf("Scenario #%d\n",z);printf("Frog Distance = %.3lf\n\n",sum);}int main(){int i,j,k;while(scanf("%d",&n),n!=0){for(i=1;i<=n;i++){scanf("%lf%lf",&s[i][1],&s[i][2]);}for(i=1;i<=n;i++){for(j=1;j<=n;j++){a[i][j]=sqrt((s[i][1]-s[j][1])*(s[i][1]-s[j][1])+(s[i][2]-s[j][2])*(s[i][2]-s[j][2]));}}l=0;for(i=1;i<=n;i++){for(j=1;j<=n;j++){if(i!=j){e[l].u=i;e[l].v=j;e[l].w=a[i][j];l++;}//printf("%.2lf    ",a[i][j]);}//printf("\n");}sort(e,e+l,cmp);/*for(i=0;i<l;i++){printf("%d %d %.2lf\n",e[i].u,e[i].v,e[i].w);}*/Kruskal();}return 0;}