hdu 3584 Cube （三维树状数组，更新区间，查询单点）

Cube

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1722    Accepted Submission(s): 898

Problem Description

Given an N*N*N cube A, whose elements are either 0 or 1. A[i, j, k] means the number in the i-th row , j-th column and k-th layer. Initially we have A[i, j, k] = 0 (1 <= i, j, k <= N). 
We define two operations, 1: “Not” operation that we change the A[i, j, k]=!A[i, j, k]. that means we change A[i, j, k] from 0->1,or 1->0. (x1<=i<=x2,y1<=j<=y2,z1<=k<=z2).
0: “Query” operation we want to get the value of A[i, j, k].

 

Input

Multi-cases.
First line contains N and M, M lines follow indicating the operation below.
Each operation contains an X, the type of operation. 1: “Not” operation and 0: “Query” operation.
If X is 1, following x1, y1, z1, x2, y2, z2.
If X is 0, following x, y, z.

 

Output

For each query output A[x, y, z] in one line. (1<=n<=100 sum of m <=10000)

 

Sample Input

2 5 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 2 2 2 0 1 1 1 0 2 2 2

 

Sample Output

1 0 1

 

Author

alpc32

 

Source

2010 ACM-ICPC Multi-University Training Contest（15）——Host by NUDT

 

三维树状数组

容斥那里注意一下。

多组数据因为忘记清空c数组而wa了1次，细心！

/*************************************************************************
    > File Name: code/hdu/3584.cpp
    > Author: 111qqz
    > Email: rkz2013@126.com 
    > Created Time: 2015年08月07日 星期五 14时01分53秒
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)  
typedef long long LL;
typedef unsigned long long ULL;
const int inf = 0x7fffffff;
const int N=1E2+5;
int c[N][N][N];
int n,m;
int x1,y1,z1,x2,y2,z2;
int lowbit( int x)
{
    return x&(-x);
}
void update ( int x,int y,int z,int delta)
{
    for ( int i = x; i <= n ; i = i + lowbit(i))
    {
    for ( int j =  y ; j <= n ; j = j + lowbit(j))
    {
        for ( int  k = z ; k  <= n ; k = k + lowbit(k))
        {
        c[i][j][k] += delta;
        }
    }
    }
}

int sum (int x,int y,int z)
{
    int res = 0;
    for ( int i = x; i >= 1 ; i -= lowbit(i))
    {
    for ( int j = y ; j >= 1 ; j -= lowbit(j))
    {
        for ( int k = z ; k >= 1 ; k -= lowbit(k))
        {
        res = res + c[i][j][k];
        }
    }
    }
    return res;
}
int main()
{
    int op;
    while (scanf("%d %d",&n,&m)!=EOF)
    {
    memset(c,0,sizeof(c));
    for ( int i = 1 ; i <= m ; i ++ )
    {
        scanf("%d",&op);
        if (op)
        {
        scanf("%d %d %d %d %d %d",&x1,&y1,&z1,&x2,&y2,&z2);
        update (x1,y1,z1,1);
        update (x1,y1,z2+1,1);
        update (x1,y2+1,z1,1);
        update (x2+1,y1,z1,1);

        update (x2+1,y2+1,z1,1);
        update (x2+1,y1,z2+1,1);
        update (x1,y2+1,z2+1,1);
        update (x2+1,y2+1,z2+1,1);

        }
        else
        {
        scanf("%d %d %d",&x1,&y1,&z1);
        cout<<sum(x1,y1,z1)%2<<endl;
        }
    }
    }
  
    return 0;
}