Cube Stacking

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

6M 1 6C 1M 2 4M 2 6C 3C 4

Sample Output

102

题目解析：

本题讲的是有n件物品，和几个操作，M代表把a所在的堆全部放到b的上方，C代表查询a上方有几个物品。原来想的模拟栈，不过一看数据我就放弃了。后来想了想，果断并查集，而且还是带权的那种。
我们需要定义两个数组，一个dis[]表示当前物品上面有多少物品，一个cnt[]表示当前物品所在的堆的物品个数，然后就是带权并查集套路。

代码如下：

  #include <iostream>      #include <cstdio>      using namespace std;      const int maxn=31000;      int father[maxn],cnt[maxn],dis[maxn];      int find(int x){          if(father[x]!=x){              int tmp=father[x];              father[x]=find(father[x]);              dis[x]+=dis[tmp];          }          return father[x];      }      void combine(int x,int y){          father[x]=y;          dis[x]+=cnt[y];          cnt[y]+=cnt[x];      }      int main(){          int m;          scanf("%d",&m);          for(int i=0;i<maxn;i++){              father[i]=i;              cnt[i]=1;              dis[i]=0;          }          while(m-- >0){              char ch;              cin>>ch;              if(ch=='M'){                  int a,b;                  scanf("%d%d",&a,&b);                  if(find(a)!=find(b)) combine(find(a),find(b));              }else{                  int x;                  scanf("%d",&x);                  find(x);                  printf("%d\n",dis[x]);              }          }          return 0;      }