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.

剛學並查集，搞了好長時間才搞明白，under陣列中每個元素代表與父節點之間的盒子個數，因為要路徑壓縮，所以需要將壓縮的點更新under值。

#include<bits/stdc++.h>
using namespace std;

const int MAX=30000;
int par[MAX+5], cnt[MAX+5], under[MAX+5];

void init(){
    for(int i=0; i<=MAX; i++){
        par[i]=i;
        cnt[i]=1;
        under[i]=0;
    }
}

int fin(int x){
    if(par[x]==x) return x;
    int temp=par[x];
    par[x]=fin(par[x]);
    under[x]+=under[temp];
    return par[x];
}

void unite(int x, int y){
    x=fin(x);
    y=fin(y);
    if(x==y) return;
    par[x]=y;
    under[x]=cnt[y];
    cnt[y]+=cnt[x];
}

int main(){
    int n;
    scanf("%d", &n);
    init();
    while(n--){
        char c;
        getchar();
        scanf("%c", &c);
        if(c=='M'){
            int a, b;
            scanf("%d%d", &a, &b);
            unite(a, b);
        }
        else{
            int a;
            scanf("%d", &a);
            fin(a);
            printf("%d\n", under[a]);
        }
    }
    return 0;
}