POJ 2253 Frogger（最小生成树）

最新推荐文章于 2017-09-21 19:42:14 发布

转载最新推荐文章于 2017-09-21 19:42:14 发布 · 90 阅读

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原文链接：http://www.cnblogs.com/cjyyb/p/7239531.html

本文解析了一个经典的青蛙跳跃问题，两只青蛙分别位于不同坐标，需要通过一系列跳跃从一只青蛙的位置到达另一只青蛙的位置。文章详细介绍了如何利用最小生成树算法来找到两青蛙间路径的最大距离最小值。

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题面

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

Sample Output

Scenario #1
Frog Distance = 5.000

Scenario #2
Frog Distance = 1.414

题解

题目大意：
有两只青蛙，所在位置分别是读入的前两个坐标
后面的坐标是其他的石头。青蛙1可以跳到任意一块石头上
问，青蛙1跳到青蛙2所在的位置的路径中，最大距离最小是多少。

题解:
这条路径显然在最小生成树上面，n^2连完边，直接求出最小生成树即可

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
#define MAX 210
int x[MAX],y[MAX],cnt,f[MAX],N;
inline int read()
{
       int x=0,t=1;char ch=getchar();
       while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
       if(ch=='-'){t=-1;ch=getchar();}
       while(ch>='0'&&ch<='9'){x=x*10+ch-48;ch=getchar();}
       return x*t;
}
inline double Dis(int u,int v)
{
       return sqrt((x[u]-x[v])*(x[u]-x[v])+(y[u]-y[v])*(y[u]-y[v]));
}
struct Line
{
       int u,v;
       double dis;
}e[50000];
bool operator <(Line a,Line b)
{
       return a.dis<b.dis;
}
int getf(int x)
{
       return x==f[x]?x:f[x]=getf(f[x]);
}
void merge(int x,int y)
{
       int a=getf(x);
       int b=getf(y);
       f[a]=b;
}
int main()
{
      int T=0;
      while(233)
      {
           N=read();
           if(!N)break;
           T++;
           for(int i=1;i<=N;++i)
              x[i]=read(),y[i]=read();
           cnt=1;
           for(int i=1;i<=N;++i)f[i]=i;
           for(int i=1;i<=N;++i)
            for(int j=1;j<=N;++j)
               if(i!=j)
                  e[cnt++]=(Line){i,j,Dis(i,j)};
           sort(&e[1],&e[cnt]);
           for(int i=1;i<=cnt;++i)
           {
                  merge(e[i].u,e[i].v);
                  if(getf(1)==getf(2))
                  {
                          printf("Scenario #%d\n",T);
                          printf("Frog Distance = %.3f\n\n",e[i].dis);
                          break;
                  }
           }
      }
      return 0;
}

转载于:https://www.cnblogs.com/cjyyb/p/7239531.html