POJ 1988 Cube Stacking（带权并查集）

Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 23678		Accepted: 8299
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

6
M 1 6
C 1
M 2 4
M 2 6
C 3
C 4

Sample Output

1
0
2

Source

USACO 2004 U S Open

题目链接：POJ 1988

以前看了很久的题解但是完全看不懂，也是最近会了食物链之后才做的。

既然食物链可以用数值代表关系，那这题只要把取模去掉就可以用数值node[x].up代表自己与祖节点之间的砖块数（不包括本身），然后还有一个问题就是还要知道自己所在砖块堆的总个数node[fx].total（fx是此砖块堆的祖节点即代表）。

那统计up用食物链的思想比较简单就可以做到，关键就是total怎么算，可以设想（画图更好理解），把砖a所在堆放到砖b所在堆，那node[fa].total肯定增加了node[fb].total个，然后node[fa].up显然不变（实际上祖节点就是堆的最上面那块砖的号码），node[fb].total就不要动了，因为只有祖节点才能具有记录total的功能，fb此时已沦为fa下面的一个子节点，node[fb].up还是可以动的，即增加了node[fa].total个。然后这题就差不多解决了，最后输出的时候用本堆砖块数total-本砖块上面的砖块数up-1，减1是因为题目要求只算压在自己这块砖下面的，自己当然不算

代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<bitset>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=300010;
struct info
{
	int pre;
	int up;
	int total;
};
info node[N];
void init(int n)
{
	for (int i=0; i<=n; ++i)
	{
		node[i].pre=i;
		node[i].up=0;
		node[i].total=1;
	}
}
int find(int n)
{
	if(node[n].pre==n)
		return n;
	int tpre=node[n].pre;
	node[n].pre=find(node[n].pre);
	node[n].up=node[n].up+node[tpre].up;
	return node[n].pre;
}
void joint(int a,int b)
{
	int fa=find(a),fb=find(b);
	if(fa!=fb)
	{
		node[fb].pre=fa;
		node[fb].up+=node[fa].total;
		node[fa].total+=node[fb].total;
	}
}
int main(void)
{
	int n,i,j,a,b,x;
	char ops[3];
	while (~scanf("%d",&n))
	{
		init(n);
		for (i=0; i<n; ++i)
		{
			scanf("%s",ops);
			if(ops[0]=='M')
			{
				scanf("%d%d",&a,&b);
				joint(a,b);
			}
			else
			{
				scanf("%d",&x);
				int fx=find(x);
				printf("%d
",node[fx].total-node[x].up-1);
			}
		}
	}
	return 0;
}