HDU 3584 树状数组

Cube

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1956 Accepted Submission(s): 1017

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

5
1 1 1  1 1 1
1 1 1
1 1 1  2 2 2
1 1 1
2 2 2

Sample Output

1
0
1

Author

alpc32

Source

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

题意：三维坐标内，每一个点初始值为0，每次改变1->0,0->1,1 111 222，表示改变（1,1,1）~（2，2,2）范围的点，0 111 表示（1,1,1）点的值几。

代码：

 1 /*
 2 这题挺巧妙，树状数组求和，因为变1次和变三次一样所以最后变得次数是奇数结果就是1，偶数结果就是0；
 3 变数的时候注意是三维的，要改变8个顶点的值才能保证求和正确。
 4 */
 5 #include<iostream>
 6 #include<cstdio>
 7 #include<cstring>
 8 using namespace std;
 9 int A[102][102][102];
10 int n,m;
11 int lowbit(int a)
12 {
13     return a&(-a);
14 }
15 void change(int a1,int b1,int c1)
16 {
17     for(int i=a1;i<=n;i+=lowbit(i))
18     {
19         for(int j=b1;j<=n;j+=lowbit(j))
20         {
21             for(int k=c1;k<=n;k+=lowbit(k))
22             {
23                 A[i][j][k]++;
24             }
25         }
26     }
27 }
28 int sum(int a1,int b1,int c1)
29 {
30     int ans=0;
31     for(int i=a1;i>0;i-=lowbit(i))
32     {
33         for(int j=b1;j>0;j-=lowbit(j))
34         {
35             for(int k=c1;k>0;k-=lowbit(k))
36             {
37                 ans+=A[i][j][k];
38             }
39         }
40     }
41     return ans&1;
42 }
43 int main()
44 {
45     int x,x1,y1,z1,x2,y2,z2;
46     while(scanf("%d%d",&n,&m)!=EOF)
47     {
48         memset(A,0,sizeof(A));
49         while(m--)
50         {
51             scanf("%d",&x);
52             if(x)
53             {
54                 scanf("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2);
55                 change(x1,y1,z1);
56                 change(x2+1,y2+1,z2+1);
57                 change(x2+1,y1,z1);
58                 change(x1,y2+1,z1);
59                 change(x1,y1,z2+1);
60                 change(x1,y2+1,z2+1);
61                 change(x2+1,y1,z2+1);
62                 change(x2+1,y2+1,z1);            
63             }
64             else
65             {
66                 scanf("%d%d%d",&x1,&y1,&z1);
67                 printf("%d
",sum(x1,y1,z1));
68             }
69         }
70     }
71     return 0;
72 }