Cube Stacking


Time Limit: 2000MS

 

Memory Limit: 30000K

Total Submissions: 18973

 

Accepted: 6605

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​​


 

题目大意：

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。

 

思路：

使用并查集，每加入两块砖头，让下边的砖头作为父节点，上边的砖头作为根节点；两堆砖头合并的时候，让下边堆的根节点作为上边堆的根节点。这其中，除了father数组外，还需要两个数组sum(用来记录每堆共有多少砖块)，under数组(用来记录第i块砖头下边有多少个砖头)，其中sum数组只在合并堆的时候更新，under数组在合并堆的时候和路径压缩的时候都要更新。

 

代码：

# include<iostream>
# include<stdio.h>
# include<string.h>
using namespace std;
const int MAXN = 30010;

int father[MAXN],sum[MAXN],under[MAXN];

int find(int a)
{
    if(a==father[a])
        return a;

    int t = find(father[a]);//找到a的根节点
    under[a] += under[father[a]];//找根节点的路径上  更新a堆下边砖块数目
    father[a] = t;//更新根节点

    return father[a];
}
void Merge(int a,int b)
//将a所在的堆叠放到b所在的堆上边
{
    int pa = find(a);
    int pb = find(b);
    if(pa==pb)
        return;
    father[pa] = pb;//将上边的堆的根结点更新为下边的堆的根节点
    under[pa] = sum[pb];
    //上边的堆增加b堆的数量，这里 = 和 += 效果一样，因为上边的堆最下边也就是a的根under值为0
    sum[pb] += sum[pa];
    //b所在堆的数目加上a堆数目
}
int main()
{
    int P, a, b;
    char cmd[20];
    for(int i = 0; i <= MAXN; i++)
    {
        father[i] = i;
        under[i] = 0;
        sum[i] = 1;
    }
    while(~scanf("%d", &P))
    {
        scanf("%s",cmd);
        if(cmd[0]=='M')
        {
            scanf("%d%d", &a, &b);
            Merge(a,b);
        }
        else
        {
            scanf("%d", &a);
            find(a);
            printf("%d\n",under[a]);
        }
    }


    return 0;
}