POJ 1988 Cube Stacking
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.
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.
* 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
6 M 1 6 C 1 M 2 4 M 2 6 C 3 C 4
Sample Output
1 0 2
并查集
有N(N<=30,000)堆方块，开始每堆都是一个
方块。方块编号1 – N. 有两种操作：
M x y ： 表示把方块x所在的堆，拿起来叠放
到y所在的堆上。
C x : 问方块x下面有多少个方块。
操作最多有 P (P<=100,000)次。对每次C操
作，输出结果。
解法：
除了parent数组，还要开设
sum数组：记录每堆一共有多少方块。
若parent[a] = a, 则sum[a]表示a所在的堆的
方块数目。
under数组， under[i]表示第i个方块下面有多少
个方块。
under数组在 堆合并和 路径压缩的时候都要更
新。View Code1 #include <iostream>
2 #include <cstring>
3 #include <cmath>
4 using namespace std;
5
6 int fa[110000],sum[110000],under[110000];
7
8 int getfather(int a)
9 {
10 if (fa[a]==a)
11 return a;
12 int t=getfather(fa[a]);
13 under[a]+=under[fa[a]];
14 fa[a]=t;
15 return t;
16 }
17
18 void merge(int a,int b)
19 {
20 int ta=getfather(a);
21 int tb=getfather(b);
22 if (ta==tb)
23 return;
24 under[ta]=sum[tb];
25 fa[ta]=tb;
26 sum[tb]+=sum[ta];
27 }
28 int main()
29 {
30 int n,a,b;
31 char ch;
32 cin>>n;
33 for (int i=1;i<=n;i++)
34 {
35 under[i]=0;
36 sum[i]=1;
37 fa[i]=i;
38 }
39 while (n--)
40 {
41 cin>>ch;
42 if (ch=='M')
43 {
44 cin>>a>>b;
45 merge(a,b);
46 }
47 else
48 {
49 cin>>a;
50 getfather(a);
51 cout <<under[a]<<endl;
52 }
53 }
54 return 0;
55 }
