http://codeforces.com/contest/707/problem/D
先说一下离线和在线：在线的意思就是每一个询问单独处理复杂度O（多少多少)，离线是指将所有的可能的询问先一次都处理出来，最后对于每个询问O(1)回答。
然后说一下cf的这题：
2 seconds
512 megabytes
standard input
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?
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.
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.
2 3 3
1 1 1
3 2
4 0
1
4
0
4 2 6
3 2
2 2 2
3 3
3 2
2 2 2
3 2
2
1
3
3
2
4
2 2 2
3 2
2 2 1
2
1
This image illustrates the second sample case.
题目大意：给你一个n*m的书架，刚开始书架上面都是空的。然后有如下四种操作
第一种：放一本书在[i][j]这个位置，如果这里有书的话，那就不用放了。
第二种：拿走[i][j]这个位置的书，如果没有书的话，那就不用拿了。
第三种：将第i行取亦或。
第四种：回到第i步操作的状态
有q个询问，输出该询问下目前书架上还有几本书？
思路：一次性全部取出来，然后建图进行dfs。所以这里的复杂度是O（q）
//看看会不会爆int!数组会不会少了一维！
//取物问题一定要小心先手胜利的条件
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define ALL(a) a.begin(), a.end()
#define pb push_back
#define mk make_pair
#define fi first
#define se second
const int maxn = + ;
const int maxq = + ;
bitset<maxn> bite[maxn];
bitset<maxn> tmp;
int n, m, q;
int a[maxq], b[maxq], op[maxq], ans[maxq];
vector<int> E[maxq]; void dfs(int u){
//printf("u = %d\n", u);
int len = E[u].size();
if (u == ){
ans[u] = ;
for (int i = ; i < len; i++){
ans[E[u][i]] = ans[u];
dfs(E[u][i]);
}
}
else if (op[u] == ){
bool flag = false;
if (!bite[a[u]][b[u]]) flag = true, bite[a[u]][b[u]] = , ans[u] += ;
for (int i = ; i < len; i++){
ans[E[u][i]] = ans[u];
dfs(E[u][i]);
}
if (flag) bite[a[u]][b[u]] = ;
}
else if (op[u] == ){
bool flag = false;
if (bite[a[u]][b[u]]) flag = true, bite[a[u]][b[u]] = , ans[u] -= ;
for (int i = ; i < len; i++){
ans[E[u][i]] = ans[u];
dfs(E[u][i]);
}
if (flag) bite[a[u]][b[u]] = ;
}
else if (op[u] == ){
ans[u] = ans[u] + m - bite[a[u]].count() - bite[a[u]].count();
bite[a[u]] ^= tmp;
for (int i = ; i < len; i++){
ans[E[u][i]] = ans[u];
dfs(E[u][i]);
}
bite[a[u]] ^= tmp;
}
else if (op[u] == ){
for (int i = ; i < len; i++){
ans[E[u][i]] = ans[u];
dfs(E[u][i]);
}
}
} int main(){
cin >> n >> m >> q;
for (int i = ; i <= m; i++)
tmp[i] = ;
for (int i = ; i <= q; i++){
scanf("%d", op + i);
if (op[i] == || op[i] == ){
scanf("%d%d", a + i, b + i);//表示摆放的位置
}
if (op[i] == ){
scanf("%d", a + i);//表示第i行取反
}
if (op[i] == ){
scanf("%d", a + i);
E[a[i]].pb(i);
}
else
E[i - ].pb(i);///让每个询问之间相连
}
dfs();
for (int i = ; i <= q; i++){
printf("%d\n", ans[i]);
}
return ;
}