（待整理）B

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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

本思路简单，但是没有路径压缩和优化，因此虽然结果正确，但是会超时，到现在也没想出怎样进行路径压缩

就是用数组记录下父节点和子节点，从而形成一个没有树枝的树，但是这样最大的坏处就是在遍历时会消耗大量的时间，不可取。

#include<iostream>
#include <cstdio>
using namespace std;
int count ;
int son[30005],fa[30005];
int find(int w)
{
 if(fa[w]==0)
  return w;
 else
 {
  count++;
 return   find(fa[w]);
 }
}
int findson(int w)
{
 if(son[w]==0)
  return w;
  else
  findson(son[w]);
}
int main()
{
 int  n,i,x,y,w;
 scanf("%d",&n); 
 memset(son,0,4*30005);
    memset(fa,0,4*30005);
 for(i=1;i<=n;i++)
 {
  count=0;
  char ch;
  getchar();
  scanf("%c",&ch);
  if(ch=='M')
  {
   scanf("%d%d",&x,&y);
   x=find(x);
   y=findson(y);
   fa[x]=y;
   son[y]=x;
  }
  else
  {
   scanf("%d",&w);
   find(w);
   printf("%d ",count);
  }  
 }
 return 0;
}

思路稍做优化，依旧超时.........但是结果依旧是正确的

#include<iostream>
#include <cstdio>
using namespace std;
int son[30005],fa[30005],count[30005];
int findfa(int x,int y)
{
 if(fa[x]==0)
 {
  count[x]=count[x]+count[y];
  return x;
 }
 else
 {
    count[x]=count[x]+count[y];
 return   findfa(fa[x],y);
 }
}
int findson(int w)
{
 if(son[w]==0)
  return w;
  else
  findson(son[w]);
}
void Count(int x,int cou )
{
 count[x]=cou+count[x];
 if(son[x]==0)
  return ;
 Count(son[x],cou);
}
int main()
{
 int  n,i,x,y,w;
 scanf("%d",&n); 
 memset(son,0,4*30005);
    memset(fa,0,4*30005);
    for( i=0;i<30005;i++)
  count[i]=1;
 for(i=1;i<=n;i++)
 {
  char ch;
  getchar();
  scanf("%c",&ch);
  if(ch=='M')
  {
   scanf("%d%d",&x,&y);
   y=findson(y);
   Count(x,count[y]);
   x=findfa(x,y);
  // cout<<"****************"<<x<<' '<<y<<endl;
   fa[x]=y;
   son[y]=x;
  }
  else
  {
   scanf("%d",&w);
   printf("%d ",count[w]-1) ;
  }  
 }
 return 0;
}

正解：

#include<cstdio>
#include<cstring>
const int maxn=100000+5;
int set[maxn],cnt[maxn],top[maxn];  //t[k]为k所在栈的栈底元素序号，top[k]为k所在栈的栈顶元素序号，cnt[k]为k到set[k]的元素个数
int set_find(int p)                     //计算元素p所在栈的栈底元素set[p]以及p到set[p]的元素个数，路径压缩
{
    if(set[p]<0)                   //栈中只有一个元素p
        return p;
    if(set[set[p]]>=0)              //的下方还有元素
    {
        int fa=set[p];
        set[p]=set_find(fa);
        cnt[p]=cnt[p]+cnt[fa];       //累加从fa到set[p]之间的元素个数,计算出每个元素到栈底的距离
    }
    return set[p];
}
void set_join(int x,int y)          //将x所在的栈移动到y所在的栈的栈顶
{
    x=set_find(x);                //将x，y分别设置为相应栈的栈底元素的序号
    y=set_find(y);i
    set[x]=y;                     //将y设置为set[x]的下一元素，两个栈连接完成
    set_find(top[y]);            //计算原先y所在栈的栈顶元素到栈底元素之间的元素个数
    cnt[x]=cnt[top[y]]+1;        //计算新栈从x到栈底的元素个数
    top[y]=top[x];               //更新y所在栈的栈顶元素为x原来所在栈的栈顶元素
}
int main()
{
    int p;
    scanf("%d",&p);              //输入操作的数目
    memset(set,-1,sizeof(set));  //将set中所有的元素都独自成栈
    memset(cnt,0,sizeof(cnt));   //清空计数
    for(int i=0; i<maxn; i++)    //初始化每一个栈的栈顶元素
        top[i]=i;
    while(p--)
    {
        char s[5];
        scanf("%s",s);               //读入当前操作
        if(s[0]=='M') 
        {
            int x,y;
            scanf("%d%d",&x,&y);   
            set_join(x,y);          //移动栈
        }
        else
        {
            int x;
            scanf("%d",&x);
            set_find(x);
            printf("%d ",cnt[x]);
        }
    }
    return 0;
}