Building Block

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1729 Accepted Submission(s): 516

Problem Description

John are playing with blocks. There are N blocks (1 <= N <= 30000) numbered 1...N。Initially, there are N piles, and each pile contains one block. Then John do some operations P times (1 <= P <= 1000000). There are two kinds of operation:

M X Y : Put the whole pile containing block X up to the pile containing Y. If X and Y are in the same pile, just ignore this command.

C X : Count the number of blocks under block X

You are request to find out the output for each C operation.

Input

The first line contains integer P. Then P lines follow, each of which contain an operation describe above.

Output

Output the count for each C operations in one line.

Sample Input

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

Sample Output

3、代码：

#include
   
    #include
    
     int num[333333];int set[333333];int under[333333];int find(int x){    int tmp;    if (x!=set[x])    {        tmp = find(set[x]);        under[x] += under[set[x]];        set[x] = tmp;    }    return set[x];}void merge(int a,int b){    int fx=find(a);    int fy=find(b);    if(fx!=fy)    {        under[fx]=num[fy];        num[fy]+=num[fx];        set[fx]=fy;    }}int main(){    int n,a,b;    char s[5];    while(scanf("%d",&n)!=EOF)    {        memset(under,0,sizeof(under));        for(int i=0; i<=n; i++)//初始化        {            set[i]=i;            num[i]=1;        }        for(int i=0; i

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