Cube Stacking

最新推荐文章于 2022-03-03 13:43:07 发布

转载最新推荐文章于 2022-03-03 13:43:07 发布 · 59 阅读

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原文链接：http://www.cnblogs.com/767355675hutaishi/p/3859066.html

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#人工智能

本文介绍了一个涉及堆叠游戏的问题，游戏中包含移动和计数两种操作。通过使用并查集的数据结构来实现高效的堆叠移动与计数，最终验证游戏的结果。代码详细展示了如何初始化并查集、查找根节点及合并两个集合。

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Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 18514		Accepted: 6426
Case Time Limit: 1000MS

Description

Farmer John and Betsy are playing a game with N (1 <= N <= 30,000)identical cubes labeled 1 through N. They start with N stacks, each containing a single cube. Farmer John asks Betsy to perform P (1<= P <= 100,000) operation. There are two types of operations:
moves and counts.
* In a move operation, Farmer John asks Bessie to move the stack containing cube X on top of the stack containing cube Y.
* In a count operation, Farmer John asks Bessie to count the number of cubes on the stack with cube X that are under the cube X and report that value.

Write a program that can verify the results of the game.

Input

* Line 1: A single integer, P

* Lines 2..P+1: Each of these lines describes a legal operation. Line 2 describes the first operation, etc. Each line begins with a 'M' for a move operation or a 'C' for a count operation. For move operations, the line also contains two integers: X and Y.For count operations, the line also contains a single integer: X.

Note that the value for N does not appear in the input file. No move operation will request a move a stack onto itself.

Output

Print the output from each of the count operations in the same order as the input file.

Sample Input

Sample Output

1
0
2

 1 #include"iostream"
 2 #include"cstdio"
 3 #include"cstring"
 4 using namespace std;
 5 const int ms=30001;
 6 int parent[ms];
 7 int sum[ms];
 8 int under[ms];
 9 int find(int x)
10 {
11     if(parent[x]==x)
12         return x;
13     int t=find(parent[x]);
14     under[x]+=under[parent[x]];
15     parent[x]=t;
16     return parent[x];
17 }
18 void merge(int a,int b)
19 {
20     int n;
21     int fa=find(a);
22     int fb=find(b);
23     if(fa==fb)
24         return ;
25     under[fb]=sum[fa];
26     parent[fb]=fa;
27     sum[fa]+=sum[fb];
28 }
29 int main()
30 {
31     int i,j,p;
32     for(i=0;i<ms;i++)
33     {
34         parent[i]=i;
35         under[i]=0;
36         sum[i]=1;
37     }
38     cin>>p;
39     char c;
40     while(p--)
41     {
42         cin>>c;
43         if(c=='M')
44         {
45             cin>>i>>j;
46             merge(j,i);
47         }
48         else
49         {
50             cin>>i;
51             find(i);
52             cout<<under[i]<<endl;
53         }
54     }
55     return 0;
56 }