Codeforces-707D：Persistent Bookcase （离线处理特殊的可持久化问题&&Bitset）

Posted 2020-10-27 ---学习ing---

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 1 to 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.

Example

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

Note

This image illustrates the second sample case.

题意：现在有一个N*M的书架，有Q个操作，对于每个操作，输入opt：

          如果opt==1，那么输入x，y，如果第x行第y列无书，则放一本书。

          如果opt==2，那么输入x，y，如果第x行第y列有书，则取走那本书。

          如果opt==3，那么输入x，将第x行有书的取走，无书的位置放一本。

          如果opt==4，那么输入k，表示把书架的情况恢复为第k次操作后的样貌，k在当前操作之前。

思路：初看可能是可持久化数据结构，但是注意到整体操作顺序为有根树，可以DFS回溯，对于书架上的情况，可以直接积累或者Bitset假设。

#include<bitset>
#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=1010;
const int maxm=100010;
bitset<maxn>s[maxn],P;
int N,M,Q;
int Laxt[maxm],Next[maxm],To[maxm],cnt;
int opt[maxm],x[maxm],y[maxm],ans[maxm];
void read(int &res)
{
    char c=getchar(); res=0;
    for(;c>‘9‘||c<‘0‘;c=getchar());
    for(;c<=‘9‘&&c>=‘0‘;c=getchar()) res=(res<<3)+(res<<1)+c-‘0‘;
}
void add(int u,int v)
{
    Next[++cnt]=Laxt[u];
    Laxt[u]=cnt;
    To[cnt]=v;
}
void dfs(int u,int Now)
{
    for(int i=Laxt[u];i;i=Next[i]){
        int v=To[i];
        if(opt[v]==1&&s[x[v]][y[v]]==0) {
                s[x[v]][y[v]]=1;
                ans[v]=Now+1;
                dfs(v,ans[v]);
                s[x[v]][y[v]]=0;
        }
        else if(opt[v]==2&&s[x[v]][y[v]]==1) {
                s[x[v]][y[v]]=0;
                ans[v]=Now-1;
                dfs(v,ans[v]);
                s[x[v]][y[v]]=1;
        }
        else if(opt[v]==3){
                ans[v]=Now-s[x[v]].count();  
                s[x[v]]^=P;  
                ans[v]+=s[x[v]].count(); 
                dfs(v,ans[v]);
                s[x[v]]^=P;
        }
        else {
                ans[v]=Now;
                dfs(v,ans[v]);
        }
     }
}
int main()
{
    read(N); read(M); read(Q);
    for(int i=1;i<=M;i++) P.set(i);
    for(int i=1;i<=Q;i++){
        scanf("%d",&opt[i]);
        if(opt[i]==1||opt[i]==2) read(x[i]),read(y[i]);
        else read(x[i]);
        if(opt[i]==4) add(x[i],i);
        else add(i-1,i);
    }
    dfs(0,0);
    for(int i=1;i<=Q;i++) printf("%d\n",ans[i]);
    return 0;
}