POJ 2253 Frogger 解题报告

分类：图论，最短路，生成树

作者： ACShiryu

时间：2011-7-28

原题：http://poj.org/problem?id=2253

Frogger

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13595		Accepted: 4521

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

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；

参考代码：

 1 #include<iostream>
 2 #include<cstdlib>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<cmath>
 7 using namespace std;
 8 pair<int ,int> a[200];            //保存n个石头的坐标
 9 double lowcost[200],closet[200];//Prim算法必备，lowcost[i]表示i距离集合的最近距离，closet[i]表示i距离集合最近的点
10 double map[200][200];            //两点之间的距离
11 int main()
12 {
13     int n;
14     int k=1;
15     while(cin>>n,n)
16     {
17         int i,j;
18         for (i = 0 ;i < n ; i++ )
19             cin>>a[i].first>>a[i].second;    //输入n个点的坐标，从0开始，也就是说题目编程求0-1的最小Frog Distance
20         memset(lowcost,0,sizeof(lowcost));    //清零
21         for ( i = 0 ; i < n ; i ++ )
22         {
23             for ( j = 0 ; j < n ; j ++ )
24             {//求任意两点的距离，保存到map中
25                 map[i][j]=1.0*sqrt(pow(1.0*abs(a[i].first-a[j].first),2)+pow(1.0*abs(a[i].second-a[j].second),2));
26             }
27         }
28         double ans=0.0;//所要求的答案，初始化为0
29         for ( i = 0 ; i< n ; i++ )
30         {//把0放入集合，则点到集合的距离此时是点到0的距离
31             lowcost[i]=map[0][i];
32             closet[i]=0;
33         }
34 
35         for ( i = 0 ; i < n - 1 ; i ++ )
36         {
37             double mindis=1.0*(1<<20);        //点到集合最小距离，初始化为最大
38             int minone;                        //到集合最小距离对应的点
39             for ( j = 0 ; j < n ; j ++ )
40             {
41                 if(lowcost[j]&&mindis>lowcost[j])
42                 {//j点不在集合中，并且j到集合的距离比最小距离还小，则更新最小距离
43                     mindis=lowcost[j];
44                     minone=j;
45                 }
46             }
47             if(ans<mindis)        //如果答案并不比更新的最小距离大
48                 ans=mindis;        //更新答案
49             lowcost[minone]=0.0;//将该点入集合
50             if(minone==1)        //如果改点是1，则水明义江找到了答案
51                 break;
52             for ( j = 0 ; j < n ; j ++ )
53             {//更新各点到集合的最小距离
54                 if(map[j][minone]<lowcost[j])
55                 {//如果minone到某点j的距离比原来的j到集合的距离要小，则更新该点到集合的距离为改点到minone的距离
56                     lowcost[j]=map[j][minone];
57                     closet[j]=minone;
58                 }
59             }
60         }
61         cout<<"Scenario #"<<k<<endl;
62         printf("Frog Distance = %.3f\n\n",ans);
63         k++;
64     }
65     return 0;
66 }