zjnu1726 STOGOVI (lca)

Description

Mirko is playing with stacks. In the beginning of the game, he has an empty stack denoted with number 0. In the

ith step of the game he will choose an existing stack denoted with v, copy it and do one of the following actions:

a. place number i on top of the new stack

b. remove the number from the top of the new stack

c. choose another stack denoted with w and count how many different numbers exist that are in the new stack

and in the stack denoted with w

The newly created stack is denoted with i.

Mirko doesn’t like to work with stacks so he wants you to write a programme that will do it for him. For each

operation of type b output the number removed from stack and for each operation of type c count the required

numbers and output how many of them there are.

Input

The first line of input contains the integer N (1 <= N <= 300000), the number of steps in Mirko’s game.

The steps of the game are chronologically denoted with the first N integers.

The ith of the following N lines contains the description of the ith step of the game in one of the following three

forms:

"a v" for operation of type a.

"b v" for operation of type b.

"c v w" for operation of type c.

The first character in the line denotes the type of operation and the following one or two denote the accompanying

stack labels that will always be integers from the interval [0,i-1].

For each operation of type b, the stack we’re removing the element from will not be empty.

Output

For each operation type b or c output the required number, each in their own line, in the order the operations

were given in the input.

Sample Input

5
a 0
a 1
b 2
c 2 3
b 4
11
a 0
a 1
a 2
a 3
a 2
c 4 5
a 5
a 6
c 8 7
b 8
b 8

Sample Output

2
1
2
2
2
8
8

Hint

In the beginning, we have the stack S0 = {}. In the first step, we copy S0 and place

number 1 on top, so S1 = {1}. In the second step, we copy S1 and place 2 on top of it, S2 = {1,2}. In the third step we

copy S2 and remove number 2 from it, S3 = {1}. In the fourth step we copy S2 and denote the copy with S4, then count

the numbers appearing in the newly created stack S4 and stack S3, the only such number is number 1 so the solution is 1.

In the fifth step we copy S4 and remove number 2 from it, S5 = {1}.

题意：一开始给你一个空栈，有3个操作。1.a v:先把编号为v的栈复制，然后在栈顶上放i 2.b v:先把编号为v的栈复制，然后去掉栈顶元素 3.c v w:先把编号为v的栈复制，然后数出同时存在于v,w栈的数的个数。开一个二叉树，如果是a操作，那么加入新的节点，如果是b操作，那么找到v的父节点，如果是c操作，那么全部输入读入后，求一个lca。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<bitset>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef long double ldb;
#define inf 99999999
#define pi acos(-1.0)
#define maxn 300005
vector<int>vec[maxn];
vector<int>::iterator it;
struct node{
    int x,y;
}c[maxn];

int jd[maxn],fa[maxn],f[maxn][25],dep[maxn],ans[maxn],daibiao[maxn],vis[maxn],siz[maxn];

int lca(int x,int y){
    int i;
    if(dep[x]<dep[y]){
        swap(x,y);
    }
    for(i=20;i>=0;i--){
        if(dep[f[x][i] ]>=dep[y]){
            x=f[x][i];
        }
    }
    if(x==y)return x;
    for(i=20;i>=0;i--){
        if(f[x][i]!=f[y][i]){
            x=f[x][i];y=f[y][i];
        }
    }
    return f[x][0];
}

int main()
{
    int n,m,i,j,x,jiedian,jiedian1,y,k;
    char s[10];
    while(scanf("%d",&m)!=EOF)
    {
        jd[1]=1;
        for(i=1;i<=m+1;i++)vec[i].clear();
        int t=1;
        daibiao[1]=1;
        dep[1]=1;
        for(i=2;i<=m+1;i++){
            scanf("%s%d",s,&x);x++;
            jiedian=jd[x];
            if(s[0]=='a'){
                t++;
                dep[t]=dep[jiedian]+1;
                jd[i]=t;
                f[t][0]=jiedian;
                daibiao[t]=i;
                c[i].x=c[i].y=-2;
            }
            else if(s[0]=='b'){
                ans[i]=daibiao[jiedian];
                jiedian1=f[jiedian][0];
                jd[i]=jiedian1;
                c[i].x=c[i].y=-1;
            }
            else if(s[0]=='c'){
                scanf("%d",&y);y++;
                jd[i]=jiedian;
                c[i].x=jd[i];c[i].y=jd[y];
            }
        }
        for(k=1;k<=20;k++){
            for(i=1;i<=m+1;i++){
                f[i][k]=f[f[i][k-1]][k-1];
            }
        }
        for(i=2;i<=m+1;i++){
            if(c[i].x==-2)continue;
            if(c[i].x==-1){
                printf("%d\n",ans[i]-1);
            }
            else{
                int gong=lca(c[i].x,c[i].y );
                printf("%d\n",dep[gong]-1);

            }
        }
    }
    return 0;
}