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

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

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：

 /*
 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；
 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。
 */
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 using namespace std;
 int A[][][];
 int n,m;
 int lowbit(int a)
 {
     return a&(-a);
 }
 void change(int a1,int b1,int c1)
 {
     for(int i=a1;i<=n;i+=lowbit(i))
     {
         for(int j=b1;j<=n;j+=lowbit(j))
         {
             for(int k=c1;k<=n;k+=lowbit(k))
             {
                 A[i][j][k]++;
             }
         }
     }
 }
 int sum(int a1,int b1,int c1)
 {
     int ans=;
     for(int i=a1;i>;i-=lowbit(i))
     {
         for(int j=b1;j>;j-=lowbit(j))
         {
             for(int k=c1;k>;k-=lowbit(k))
             {
                 ans+=A[i][j][k];
             }
         }
     }
     return ans&;
 }
 int main()
 {
     int x,x1,y1,z1,x2,y2,z2;
     while(scanf("%d%d",&n,&m)!=EOF)
     {
         memset(A,,sizeof(A));
         while(m--)
         {
             scanf("%d",&x);
             if(x)
             {
                 scanf("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2);
                 change(x1,y1,z1);
                 change(x2+,y2+,z2+);
                 change(x2+,y1,z1);
                 change(x1,y2+,z1);
                 change(x1,y1,z2+);
                 change(x1,y2+,z2+);
                 change(x2+,y1,z2+);
                 change(x2+,y2+,z1);
             }
             else
             {
                 scanf("%d%d%d",&x1,&y1,&z1);
                 printf("%d\n",sum(x1,y1,z1));
             }
         }
     }
     return ;
 }