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

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
      
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                  #define y0 abc111qqz#define y1 hust111qqz#define yn hez111qqz#define j1 cute111qqz#define tm crazy111qqz#define lr dying111qqzusing namespace std;#define REP(i, n) for (int i=0;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<
                   
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转载于:https://www.cnblogs.com/111qqz/p/4710660.html

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