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
Sample Output
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}.
題意：一開始給你一個空棧，有3個操作。1.a v:先把編號爲v的棧複製，然後在棧頂上放i 2.b v:先把編號爲v的棧複製，然後去掉棧頂元素 3.c v w:先把編號爲v的棧複製，然後數出同時存在於v,w棧的數的個數。開一個二叉樹，如果是a操作，那麼加入新的節點，如果是b操作，那麼找到v的父節點，如果是c操作，那麼全部輸入讀入後，求一個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;
}