POJ2253 Frogger

Frogger

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 17471		Accepted: 5692

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

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

Source

Ulm Local 1997

 

 

思路：首先使用KRUSKAL求出最小生成树，然后再使用DFS遍历刚刚求出的最小生成树，找出从开始点到结束点的路径，找出这条路径上的最大长度的边。

 

#include <cstdlib>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>


#define MAXDOUBLE 1.0e30
using namespace std;


struct Pos{
       double x,y;
       }pos[200+4];//存储各个点的坐标 


struct Edge{
       int v1,v2;
       double cost;
       }edge[200*200];//使用kruskal的数据结构 

int len=0;



double g[200+1][200+1];//存储kruskal求出的那些边 


int fa[200+4];

int findfa(int v)
{
    if(fa[v]!=v)
    {fa[v]=findfa(fa[v]);}
    return fa[v];
}




bool cmp(Edge x,Edge y)
{
     if(x.cost<y.cost)
     return true;
     return false;
}

int vis[200+1];//用于DFS时起标记作用 

int dfs(int v,int n,double &mincost)
{
    int i,j,k;
    
    double tmp;
    
    if(v==1)
    return 1;
    
    for(i=1;i<n;i++)
    {if((vis[i]==0)&&(g[v][i]!=-1))
     {vis[i]=1;
     tmp=mincost;
     
     if(mincost<g[v][i])
     mincost=g[v][i];
     
     
     
     if(1==dfs(i,n,mincost)) 
     {return 1;
     
     }
     mincost=tmp;
     
    }
   }
    
    return 0;
}



    
    
    
    
int main(int argc, char *argv[])
{   
   
    int n;
    int i,j,k;
    
    int count=0;
    
    while(scanf("%d",&n)!=EOF)
    {
                              if(n==0)
                              break;
                              
                              count++;
                              
                              for(i=0;i<n;i++)
                              {cin>>pos[i].x>>pos[i].y;}
                              
                              len=0;
                              
                              
                              for(i=0;i<n;i++)
                              for(j=0;j<i;j++)
                              {
                                              edge[len].v1=i;
                                              edge[len].v2=j;
                                              edge[len].cost=sqrt((pos[i].x-pos[j].x)*(pos[i].x-pos[j].x)+(pos[i].y-pos[j].y)*(pos[i].y-pos[j].y));
                                              len++;
                              }
                              
                              
                              sort(edge,edge+len,cmp);
                              
                              
                              for(i=0;i<n;i++)
                              {fa[i]=i;}
                              
                              
                              int num=0;
                              for(i=0;i<n;i++)
                              for(j=0;j<n;j++)
                              {g[i][j]=-1;}
                              
                              
                              for(i=0;i<len;i++)
                              {
                                              int fv1=findfa(edge[i].v1);
                                              int fv2=findfa(edge[i].v2);
                                              if(fv1!=fv2)
                                              {fa[fv1]=fv2;
                                               g[edge[i].v1][edge[i].v2]=edge[i].cost;
                                               g[edge[i].v2][edge[i].v1]=edge[i].cost;
                                              }
                                              
                              }
                              
                              
                              
                              
                              for(i=0;i<n;i++)
                              {vis[i]=0;}
                              
                              double maxmincost=0;
                              
                              dfs(0,n,maxmincost);
                              float maxmincost2=(float)maxmincost;
                              printf("Scenario #%d\nFrog Distance = %.3f\n\n",count,maxmincost2);
    }

       

                              
                                               
                                               
                                               
                                               
                                               
                              
                              
                              
                              
                              
                              
                              
                                              
                                              
                              
                              
                              
                              
                              
                              
                              
    
   
                  
                  
                  
    
    //system("PAUSE");
    return EXIT_SUCCESS;
}