poj 1988 Cube Stacking（带权并查集）

Time Limit: 2000MS Memory Limit: 30000KTotal Submissions: 22247 Accepted: 7807Case 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

6M 1 6C 1M 2 4M 2 6C 3C 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);      }    }  }}