Codeforces Round #368 (Div. 2) D 离线+查询树

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本文解析了Codeforces平台上编号为707的问题D，通过离线构建查询树的方法解决了一个涉及书架状态更新的问题。具体介绍了四种操作及其处理方式，并提供了完整的C++代码实现。

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题目传送门： http://codeforces.com/contest/707/problem/D

题意：有一个书架，书架有n层，每层有m个位置，接下来是q组操作，操作有四种方式，输出每次操作后书架中书的数量

四种操作：

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.

思路：离线建一个查询树即可

代码如下：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <stdio.h>
#include <string>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <stack>
#include <bitset>

using namespace std;
#define   lson          l,m,rt<<1
#define   rson          m+1,r,rt<<1|1
#define   ll            long long
#define   ull           unsigned long long
#define   mem(n,v)      memset(n,v,sizeof(n))
#define   MAX           100005
#define   MAXN          200005
#define   PI            3.1415926
#define   E             2.718281828459
#define   opnin         freopen("text.in.txt","r",stdin)
#define   opnout        freopen("text.out.txt","w",stdout)
#define   clsin         fclose(stdin)
#define   clsout        fclose(stdout)
#define   haha1         cout << "haha1"<< endl
#define   haha2         cout << "haha2"<< endl
#define   haha3         cout << "haha3"<< endl

const int    INF    =   0x3f3f3f3f;
const ll     INFF   =   0x3f3f3f3f3f3f3f3f;
const double pi     =   3.141592653589793;
const double inf    =   1e18;
const double eps    =   1e-8;
const ll     mod    =   1e9+7;
const ull    mx     =   133333331;
/**************************************************************************/


bitset <1005> book[1005];
int n,m,q;
int head[MAX];
int cnt;
int num;
struct edge
{
    int val,to,next;
    int x,y;
    edge(int v = 0,int t = 0,int n = -1,int a = 0,int b = 0) : val(v),to(t),next(n),x(a),y(b) {}
}edges[MAXN];
int ans[MAX];
int vis[MAX];
void init()
{
    mem(vis,0);
    mem(book,0);
    mem(ans,0);
    mem(head,-1);
    mem(edges,0);
    cnt = 0;
    num = 0;
}
void addedge(int u,int v,int val,int x,int y)
{
    edges[cnt].to = v;
    edges[cnt].val = val;
    edges[cnt].next = head[u];
    edges[cnt].x = x;
    edges[cnt].y = y;
    head[u] = cnt++;

}
void dfs(int u)
{
    vis[u] = 1;
    for(int i=head[u];i != -1;i = edges[i].next){
        int t = edges[i].to;
        int v = edges[i].val;
        int x = edges[i].x;
        int y = edges[i].y;
        if(v == 1){
            if(!book[x][y]){
                num++;
                book[x][y] = 1;
                ans[t] = num;
                if(!vis[t]) dfs(t);
                book[x][y] = 0;
                num--;
            }
            else{
                ans[t] = num;
                if(!vis[t]) dfs(t);
            }
        }
        else if(v == 2){
            if(book[x][y]){
                num--;
                book[x][y] = 0;
                ans[t] = num;
                if(!vis[t]) dfs(t);
                book[x][y] = 1;
                num++;
            }
            else{
                ans[t] = num;
                if(!vis[t]) dfs(t);
            }
        }
        else if(v == 3){
            for(int i=1;i<=m;i++){
                if(!book[x][i]) num++;
                else num--;
                book[x][i] = 1 - book[x][i];
            }
            ans[t] = num;
            if(!vis[t]) dfs(t);
            for(int i=1;i<=m;i++){
                if(!book[x][i]) num++;
                else num--;
                book[x][i] = 1 - book[x][i];
            }
        }
        else{
            ans[t] = num;
            if(!vis[t]) dfs(t);
        }
    }
}
int main()
{
    init();
    cin >> n >> m >> q;
    int id = 0;
    for(int i=1;i<=q;i++){
        int val = 0,x = 0,y = 0;
        scanf("%d",&val);
        if(val == 1 || val == 2) scanf("%d %d",&x,&y);
        else scanf("%d",&x);
        if(val == 4) id = x;
        addedge(id,i,val,x,y);
        if(val != 4) id = i;
    }
    dfs(0);
    for(int i=1;i<=q;i++)
        printf("%d\n",ans[i]);
}