三维树状数组（区间更新，单点查询）POJ

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

      树状数组学了好久了，也做了好几个题，今天想起来写博客才发现是第一个，看来之前是疏乎了。

      这是一道三维的树状数组，刚看起来很难的，不过说起来几维都是一样的，不过是个循环多少层的问题罢了。在区间更新的时候，要注意的是一维是直接减；二维是减两部分再加上那部分重复减掉的，画一个图，小学生就可以理解的道理；三维自然也是要减完再加上重复减掉的地方的，但是这里就抽象了很多，我找到这个关键，是画了三遍正方体后才成功的，不过画的难看，也不会贴图，所以就不上图了。至于这个规律在代码里体现出来，其实看一看大概就能明白的。

代码如下：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
int c[110][110][110];
int N,M;
inline int lowBit(int x)
{
return x&(-x);
}
void update(int x,int y,int z,int k)
{
for(int i=x; i<=N; i+=lowBit(i))//三维顺序
{
for(int j=y; j<=N; j+=lowBit(j))
{
for(int l=z; l<=N; l+=lowBit(l))
{
c[i][j][l]+=k;
}
}
}
}
int getsum(int x,int y,int z)
{
int sum=0;
for(int i=x; i>0; i-=lowBit(i))
{
for(int j=y; j>0; j-=lowBit(j))
{
for(int k=z; k>0; k-=lowBit(k))
{
sum+=c[i][j][k];
}
}
}
return sum;
}
void UPdate(int x1,int y1,int z1,int x2,int y2,int z2)
{                                           //三维的区间更新

update(x2+1,y2+1,z2+1,1);//最大块
update(x1,y1,z1,5);      //最小块
update(x1,y1,z2+1,-1);//其他块
update(x1,y2+1,z1,-1);
update(x2+1,y1,z1,-1);
update(x1,y2+1,z2+1,-1);
update(x2+1,y1,z2+1,-1);
update(x2+1,y2+1,z1,-1);

}
int main()
{
int x1,x2,y1,y2,z1,z2;
int k,ans;
while(~scanf("%d%d",&N,&M) && N && M)
{
memset(c,0,sizeof(c));
for(int i=1; i<=M; i++)
{
scanf("%d%d%d%d",&k,&x1,&y1,&z1);
if(k==0)    //单点查询
{
ans=getsum(x1,y1,z1);
printf("%d\n",ans%2);
}
else        //区间更新
{
scanf("%d%d%d",&x2,&y2,&z2);
UPdate(x1,y1,z1,x2,y2,z2);
}
}
}
return 0;
}