Cube Stacking

Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
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

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

Sample Output

1
0
2
分析：题意是支持2种操作，1是把包含x的联通块移到y上，2是询问x下面有几个物品；
　　　带权并查集，p[x]为x的根，cnt[x]为x联通块大小，ret[x]为答案；
　　　merge(i,j)时，先找到根x，y,p[x]=y，表示x移到y，ret[x]+=cnt[y],更新x答案，同时注意更新cnt[y]+=cnt[x]；
　　　剩下路径压缩即可；
代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <bitset>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define sys system("pause")
const int maxn=1e5+10;
const int N=5e4+10;
const int M=N*10*10;
using namespace std;
inline ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
inline ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
inline void umax(ll &p,ll q){if(p<q)p=q;}
inline void umin(ll &p,ll q){if(p>q)p=q;}
int n,m,k,t,p[maxn],ret[maxn],cnt[maxn];
char op[2];
int find(int x)
{
    if(x==p[x])return x;
    int y=p[x];
    p[x]=find(p[x]);
    ret[x]+=ret[y];
    return p[x];
}
int main()
{
    int i,j;
    scanf("%d",&n);
    rep(i,1,n)p[i]=i,cnt[i]=1;
    rep(i,1,n)
    {
        scanf("%s%d",op,&j);
        if(op[0]=='M')
        {
            scanf("%d",&k);
            int x=find(j),y=find(k);
            p[x]=y;
            ret[x]+=cnt[y];
            cnt[y]+=cnt[x];
        }
        else
        {
            find(j);
            printf("%d
",ret[j]);
        }
    }
    return 0;
}