并查集+路径压缩(poj1988)

http://poj.org/problem?id=1988

Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 19122		Accepted: 6664
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

题意：可以把题目中的意思抽象为栈集合的合并，题目中有两个操作

M 把包含x的栈放在包含y的栈的上面；

C统计包含x的栈中x下面有多少个元素；

分析：此题是并查集路径压缩的典型题目，与前面的种类并查集还有些不同；题目中用到f[],num[],dis[]三个数组，f数组记录i的父节点，num数组记录以i为根的集合有多少个元素，dis记录i到根节点的距离，那么i下面的元素就是x=finde(i)；num[x]-dis[i]-1个元素;

程序：

#include"stdio.h"
#include"string.h"
#include"iostream"
#include"map"
#include"string"
#include"queue"
#include"stdlib.h"
#include"math.h"
#define M 30009
#define eps 1e-10
#define inf 1000000000
#define mod 2333333
using namespace std;
int f[M],num[M],dis[M];
int finde(int x)
{
    if(x!=f[x])
    {
        int t=f[x];
        f[x]=finde(f[x]);
        dis[x]+=dis[t];
    }
    return f[x];
}
void make(int a,int b)
{
    int x=finde(a);
    int y=finde(b);
    if(x!=y)
    {
        f[y]=x;
        dis[y]+=num[x];
        num[x]+=num[y];
    }
}
int main()
{
    int n,i,a,b;
    char str[3];
    while(scanf("%d",&n)!=-1)
    {
        for(i=1;i<=M+1;i++)
        {
            f[i]=i;
            num[i]=1;
            dis[i]=0;
        }
        while(n--)
        {
            scanf("%s",str);
            if(str[0]=='M')
            {
                scanf("%d%d",&a,&b);
                make(a,b);
            }
            else
            {
                scanf("%d",&a);
                int x=finde(a);
                printf("%d\n",num[x]-dis[a]-1);
            }
        }
    }
}