POJ1988 Cube Stacking [并查集]

                                                                                                                      Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 15619		Accepted: 5292
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

题目大意：初始时有n个栈,每个栈中只有一个元素,按顺序标为1-n.给出两种操作M,C.M x y:把含有x的栈移至含y的栈后面.C x:查询x所在栈中在x前面的元素个数.

此题可以用并查集，用tree[]数组表示各个数的父亲节点，属于同一集合的在同一栈中，对根节点x，tree[x]<0;-tree[x]表示整个集合的元素个数。而num[]数组表示在同一集合中，子节点到父亲节点的物理元素个数。通过find，路径压缩操作就可得到当前元素到根节点的数目。

代码：

#include<iostream>
#include<cstdio>
#define N 30005
using namespace std;
int p;
int tree[N],num[N];
void init()
{
    for(int i=0;i<N;i++)
     {
         tree[i]=-1;
         num[i]=0;
     }
}
int find(int x)
{
    if(tree[x]<=0) return x;
    int t=tree[x];
    tree[x]=find(tree[x]);
    num[x]=num[x]+num[t];
    return tree[x];
}
void union_set(int a,int b)
{
    int x,y;
    x=find(a);
    y=find(b);
    if(x==y) return ;
    int temp=tree[x];
    tree[x]=y;
    num[x]=-tree[y];
    tree[y]=tree[y]+temp;
}
int main()
{
   // freopen("in.txt","r",stdin);
    char c;
    int a,b;
    scanf("%d",&p);
    getchar();
    init();
    while(p--)
    {
        scanf("%c",&c);
        if(c=='M')
        {
            scanf("%d%d",&a,&b);
            getchar();
            union_set(a,b);
        }
        else if(c=='C')
        {
            scanf("%d",&a);
            getchar();
            if(find(a)==a)
            printf("0\n");
           else
            printf("%d\n",num[a]);
        }
    }
    return 0;
}