POJ 1988 Cube Stacking

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

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【题意】

有几个stack，初始里面有一个cube。支持两种操作：1.move x y： 将x所在的stack移动到y所在stack的顶部。2.count x：数在x所在stack中，在x之下的cube的个数。

【题解】

并查集。与银河英雄传说类似。

维护两个量s[x]和d[x]，表示子树大小和到根的距离。

---

#include<iostream>
#include<cstdio>
using namespace std;
const int maxn=30000;
int p;
struct cb
{
    int fa,num,sz;
}a[maxn+5];
int fnd(int x)
{
    if(a[x].fa!=x)
    {
        int ff=a[x].fa;
        a[x].fa=fnd(ff);
        a[x].num+=a[ff].num-1;
    }
    return a[x].fa;
}
int main()
{
    for(int i=1;i<=maxn;i++)
        a[i].fa=i,a[i].sz=1,a[i].num=1;
    scanf("%d",&p);
    char ch[11];
    int x,y;
    while(p--)
    {
        scanf("%s",ch);
        if(ch[0]=='M')
        {
            scanf("%d%d",&x,&y);
            int fx=fnd(x),fy=fnd(y);
            a[fx].num+=a[fy].sz;
            a[fy].sz+=a[fx].sz;
            a[fx].fa=fy;
        }
        else
        {
            scanf("%d",&x);
            fnd(x);
            printf("%d\n",a[x].num-1);
        }
    }
    return 0;
}