PKU POJ 百练 2253 Frogger

思路 用Floyd算法（弗洛伊德）算法

http://zh.wikipedia.org/wiki/Floyd

http://baike.baidu.com/view/14495.htm

题目

时间限制:

1000ms

内存限制:

65536kB

描述

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.

输入

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.

输出

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.

样例输入

20 03 4317 419 418 50

样例输出

Scenario #1Frog Distance = 5.000Scenario #2Frog Distance = 1.414

大意

题目大意，有两只青蛙，分别在两个石头上，青蛙A想要到青蛙B那儿去，他可以直接跳到B的石头上，也可以跳到其他石头上，再从其他石头跳到B那儿，求青蛙从A到B的所有路径中最小的Frog Distance，我们定义Frog Distance为从A到B的一条路径中所跳的最大距离，例如，如果从A到B某条路径跳的距离是2，5，6，4，则Frog Distance就是6，题目输入的第一行代表石头的个数，当个数为0时结束程序，接着有n行，其中第2，3行分别代表A，B青蛙的坐标，其他n-2行分别代表空的石头的坐标，输出一个小数（保留三位），具体格式参见样例，注意没输出一个答案还要再空一行。
题目数据1很明显为5.000
对于数据2青蛙有两种方案
方案1：1-2则经过距离为2.000故此时Frog Distance=2.000
方案2：1-3-2 则经过距离分别是1.414 1.414 故此时Frog Distance=1.414
故所求的最小的Frog Distance=1.414
这道题和POJ1797比较类似，那个是求最大生成树的最小权，这个是求最小生成树的最大权，哪题是用Kruskal+并查集做的，比较麻烦，则此从网上搜了小Prim算法，果然比较方面，开始时从图中取出点0（数组从0开始），入集合，然后搜索集合外的点到集合的距离，找出距离最小的点，入集合，重复该步骤，直到点1也进入了集合，则此时的权值就是所求的值。
刚开始输出没注意，WA了一次，这还是要提醒我们要小心注意题目的输入输出，别遗漏，确保万无一失才能交；

 

#include<stdio.h>
#include<math.h>
typedef struct Point{
 int x,y;
}Point;
int main(){
 int i,j,k,n,cases=0;
 Point point[201];
 int d[201][201];
 while(scanf("%d",&n)!=EOF&&n){
  cases++;
  for(i=1;i<=n;i++)
   for(j=1;j<=n;j++)
    d[i][j]=0;
  for(i=1;i<=n;i++)
   scanf("%d%d",&point[i].x,&point[i].y);
  for(i=1;i<=n-1;i++)
   for(j=i+1;j<=n;j++)
    d[i][j]=d[j][i]=(point[i].x-point[j].x)*(point[i].x-point[j].x)+(point[i].y-point[j].y)*(point[i].y-point[j].y);
  for(k=1;k<=n;k++)
   for(i=1;i<=n-1;i++)
    for(j=i+1;j<=n;j++)
     if((d[i][k]<d[i][j])&&(d[k][j]<d[i][j])&&(d[i][k])&&(d[k][j]))//当边ik,kj的权值都小于ij时，则走i->k->j路线，否则走i->j路线
      if(d[i][k]<d[k][j]) //当走i->k->j路线时，选择max{ik,kj},只有选择最大跳才能保证连通
       d[i][j]=d[j][i]=d[k][j];
      else
       d[i][j]=d[j][i]=d[i][k];
  printf("Scenario #%d\nFrog Distance = %0.3f\n\n",cases,sqrt((float)d[1][2]));
  }
 return 0;
}