codeforces 707D D. Persistent Bookcase(dfs)

题目链接：

D. Persistent Bookcase

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

Recently in school Alina has learned what are the persistent data structures: they are data structures that always preserves the previous version of itself and access to it when it is modified.

After reaching home Alina decided to invent her own persistent data structure. Inventing didn't take long: there is a bookcase right behind her bed. Alina thinks that the bookcase is a good choice for a persistent data structure. Initially the bookcase is empty, thus there is no book at any position at any shelf.

The bookcase consists of n shelves, and each shelf has exactly m positions for books at it. Alina enumerates shelves by integers from 1to n and positions at shelves — from 1 to m. Initially the bookcase is empty, thus there is no book at any position at any shelf in it.

Alina wrote down q operations, which will be consecutively applied to the bookcase. Each of the operations has one of four types:

• 1 i j — Place a book at position j at shelf i if there is no book at it.
• 2 i j — Remove the book from position j at shelf i if there is a book at it.
• 3 i — Invert book placing at shelf i. This means that from every position at shelf i which has a book at it, the book should be removed, and at every position at shelf i which has not book at it, a book should be placed.
• 4 k — Return the books in the bookcase in a state they were after applying k-th operation. In particular, k = 0 means that the bookcase should be in initial state, thus every book in the bookcase should be removed from its position.

After applying each of operation Alina is interested in the number of books in the bookcase. Alina got 'A' in the school and had no problem finding this values. Will you do so?

Input

The first line of the input contains three integers n, m and q (1 ≤ n, m ≤ 103, 1 ≤ q ≤ 105) — the bookcase dimensions and the number of operations respectively.

The next q lines describes operations in chronological order — i-th of them describes i-th operation in one of the four formats described in the statement.

It is guaranteed that shelf indices and position indices are correct, and in each of fourth-type operation the number k corresponds to some operation before it or equals to 0.

Output

For each operation, print the number of books in the bookcase after applying it in a separate line. The answers should be printed in chronological order.

Examples

input

2 3 3
1 1 1
3 2
4 0

output

1
4
0

input

4 2 6
3 2
2 2 2
3 3
3 2
2 2 2
3 2

output

2
1
3
3
2
4

input

2 2 2
3 2
2 2 1

output

2
1

题意：

给四种操作,1是使位置(i,j)有一本书,2是使位置(i,j)为空,3是使第i行反转,4是使当前状态转变成第i步操作后的状态;每次操作输出操作完后有多少本书;

思路：

可以发现如果是1,2,3那么第i个状态就是第i-1个状态转移过来的,如果是4操作,那么就这状态就由x[i]转移过来,这样就可以得到的是一棵有根树,然后一遍dfs遍历
1,2,3操作都是可逆的,dfs在往下走的时候和往回走的时候1和2操作相反,3和3相反,所以就很好得到答案了;

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
 
using namespace std;
 
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
 
typedef  long long LL;
 
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}
 
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e5+10;
const int maxn=1e3+20;
const double eps=1e-12;

int n,m,q;
int mp[maxn][maxn],ans[N],sum=0,op[N],x[N],y[N],flag[N];
vector<int>ve[N];

int dfs(int cur)
{
    if(op[cur]==1)
    {
        if(!mp[x[cur]][y[cur]])flag[cur]=1,mp[x[cur]][y[cur]]=1,sum++;
        ans[cur]=sum;
    }
    else if(op[cur]==2)
    {
        if(mp[x[cur]][y[cur]])flag[cur]=1,mp[x[cur]][y[cur]]=0,sum--;
        ans[cur]=sum;
    }
    else if(op[cur]==3)
    {
        for(int i=1;i<=m;i++)
        {
            if(mp[x[cur]][i])mp[x[cur]][i]=0,sum--;
            else mp[x[cur]][i]=1,sum++;
        }
        ans[cur]=sum;
    }
    else if(op[cur]==4)
    {
        ans[cur]=ans[x[cur]];
    }
    int len=ve[cur].size();
    for(int i=0;i<len;i++)dfs(ve[cur][i]);


    if(op[cur]==2)
    {
        if(!mp[x[cur]][y[cur]]&&flag[cur])mp[x[cur]][y[cur]]=1,sum++;
    }
    else if(op[cur]==1)
    {
        if(mp[x[cur]][y[cur]]&&flag[cur])mp[x[cur]][y[cur]]=0,sum--;
    }
    else if(op[cur]==3)
    {
        for(int i=1;i<=m;i++)
        {
            if(mp[x[cur]][i])mp[x[cur]][i]=0,sum--;
            else mp[x[cur]][i]=1,sum++;
        }
    }
}
int main()
{
    read(n);read(m);read(q);
    op[0]=0;
    for(int i=1;i<=q;i++)
    {
        read(op[i]);read(x[i]);
        if(op[i]==4){ve[x[i]].push_back(i);continue;}
        else if(op[i]==1||op[i]==2)read(y[i]);
        ve[i-1].push_back(i);
    }
    dfs(0);
    for(int i=1;i<=q;i++)print(ans[i]);

    return 0;
}