Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 18834		Accepted: 6535
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

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

Sample Output

102

Source

题目大意：

有N个立方体和N个格子，1~N编号，一开始i立方体在i号格子上，每个格子刚好1个立方体。现在m组操作，M a b表示将a号立方体所在的格子的全部立方体放在b号立方体所在的格子的全部立方体上面。C x表示询问x号立方体下面的立方体的个数。

解题思路：

在并查集的基础上，只需要知道x到父亲的距离以及父亲到底的距离就知道x到底的距离。

解题代码：

#include 
    
     #include 
     
      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
      
       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;}