Cube Stacking

Cube Stacking
Time Limit: 2000MS Memory Limit: 30000K
Total Submissions: 21350 Accepted: 7470
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
并查集的合并,每个点记录到栈底的距离,栈底的元素记录栈的大小方便合并

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define LL long long
using namespace std;

const int MAX =  31000;

const int INF = 0x3f3f3f3f;

struct node
{
    int pre;
    int dis;
    int Size;
}a[MAX];

int m;

int Find(int x)
{
    if(a[x].pre==x)
    {
        return x;
    }
    int t=a[x].pre;
    a[x].pre=Find(t);
    a[x].dis+=a[t].dis;//将距离更新
    return a[x].pre;
}

void Join(int x,int y)
{
    int b=Find(x);
    int c=Find(y);
    a[b].pre=c;//将两个栈的栈底相连
    a[b].dis+=a[c].Size;//上面的栈底更新距离
    a[c].Size+=a[b].Size;//下面的栈底更新大小
    a[b].Size=0;//不在为栈底,大小为零
}

int main()
{
    scanf("%d",&m);
    char s[5];
    int u,v;
    for(int i=0;i<MAX;i++)
    {
        a[i].dis=0;
        a[i].Size=1;
        a[i].pre=i;
    }
    for(int i=0;i<m;i++)
    {
        scanf("%s",s);
        if(s[0]=='M')
        {
            scanf("%d %d",&u,&v);
            Join(u,v);
        }
        else
        {
            scanf("%d",&u);
            Find(u);//要先更新一遍
            printf("%d\n",a[u].dis);
        }
    }
    return 0;
}