poj 1988 Cube Stacking（带权并查集）

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 22247		Accepted: 7807
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

USACO 2004 U S Open

题目大意：

有n个立方体n个栈，一个栈一个立方体，一条线放置，标号与位置相同，第i个标号为i，现在有两类操作：M x y把x所在的立方体栈放在y所在的上面，C x，计算x下面有多少个立方体

解题思路：

rank[i]表示i所在的集合里有多少个立方体，（只用到根节点rank）。dist[i]表示i到当前栈顶的距离。那么只需要让栈顶做根节点，合并的时候把y合到x上，y栈顶的dist+=x集合的rank（注意只有虽然所有y的子节点距离都增加了，但是这里只更新了y），x栈顶的rank加上y栈顶的rank，就完成了维护。询问C x的时候就是x所在集合的rank-dist[x]-1（dist[x]在求rank的时候用到路径压缩，加上了所有祖先的dist，有种lazy的感觉。。。）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn = 30000+10;
int fa[maxn];
int rank[maxn];
int dist[maxn];
void init(){
for(int i = 0; i < maxn; i++)
{
fa[i] = i;
rank[i] = 1;
dist[i] = 0;
}
}
int find(int x)
{
if(x != fa[x])
{
int t = fa[x];
fa[x] = find(fa[x]);
dist[x] += dist[t];
}
return fa[x];
}
void unite(int x,int y)
{
int faA = find(x);
int faB = find(y);
if(faA != faB)
{
fa[faB] = faA;
dist[faB] = rank[faA];
rank[faA] += rank[faB];
}
}

int main()
{
int n;
while(~scanf("%d",&n))
{
init();
char op;
while(n--)
{
cin >> op;
int a,b;
if(op=='M')
{
scanf("%d%d",&a,&b);
unite(a,b);
}
else
{
scanf("%d",&a);
int x = find(a);
printf("%d\n",rank[x]-dist[a]-1);
}
}
}
}