POJ1988 Cube Stacking【带权并查集统计】

Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 28711		Accepted: 10081
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

Source

USACO 2004 U S Open

问题链接：POJ1988 Cube Stacking

问题描述：有n个相同的立方体（从1开始编号），开始有n堆（每堆一个），有P次操作，每次有两种选择，操作一M X Y，将包含立方体X的一堆移动包含立方体Y的一堆的上面，操作二C X，查询立方体X下面的立方体的个数。对于每个查询操作，输出相应的答案。

解题思路:带权并查集，让每堆中各个立方体都以最顶上的立方体为祖先结点，设结点 i 的祖先结点为 j ，cnt[i] 表示结点 i 到 j之间的立方体数，sum[j]表示以结点 j 为祖先结点的立方体数，则C i 的结果为 sum[j] - cnt[i]

AC的C++代码：

#include<iostream>

using namespace std;

const int N=30010;
int pre[N],cnt[N],sum[N];//pre存储结点i的父节点，cnt存储结点i到父结点之间的立方体数，sum存储以结点i为祖先结点的立方体数

void init()
{
	for(int i=0;i<N;i++){
		pre[i]=i;//初始化每个结点的父结点为自己 
		cnt[i]=0;//自己到自己之间只有0个 
		sum[i]=1;//本身只有一个 
	}
} 

int find(int x)
{
	if(x!=pre[x]){
		int r=pre[x];
		pre[x]=find(r);
		cnt[x]+=cnt[r];//原来x的父结点为r，x到r之间的立方体数为cnt[x]。现在x的父结点为pre[x],x到pre[x]之间的立方体数为cnt[i]+=cnt[r] 
	}
	return pre[x];
}

//将x这一堆移动到y这一堆的顶上 
void join(int x,int y)
{
	int fx=find(x);//x一堆的顶上立方体为fx 
	int fy=find(y);//y一堆的顶上立方体为fy 
	if(fx!=fy){
		pre[fy]=fx;//由于是x这一堆移动到y这一堆上，因此fy的父节点为fx 
		cnt[fy]=sum[fx];//fy到父节点fx之间的立方体数=原来x这一堆的立方体数 （即sum[fx]） 
		sum[fx]+=sum[fy];//如今以fx为根结点的立方体数是两堆立方体数之和 
	}
}

int main()
{
	int p;
	init();
	scanf("%d",&p);
	while(p--){
		char c;
		int x,y;
		scanf(" %c",&c);
		if(c=='M'){
			scanf("%d%d",&x,&y);
			join(x,y);
		}
		else if(c=='C'){
			scanf("%d",&x);
			int r=find(x);
			printf("%d\n",sum[r]-cnt[x]-1);
		}	
	}
	return 0;
}

POJ1988 Cube Stacking【带权并查集 统计】

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POJ1988 Cube Stacking【带权并查集统计】