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


[align=left]Problem Description[/align]
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].

[align=left]Input[/align]
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.

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

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

1
0
1

[align=left]Author[/align]
alpc32

[align=left]Source[/align]
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[102][102][102];
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=0;
for(int i=a1;i>0;i-=lowbit(i))
{
for(int j=b1;j>0;j-=lowbit(j))
{
for(int k=c1;k>0;k-=lowbit(k))
{
ans+=A[i][j][k];
}
}
}
return ans&1;
}
int main()
{
int x,x1,y1,z1,x2,y2,z2;
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(A,0,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+1,y2+1,z2+1);
change(x2+1,y1,z1);
change(x1,y2+1,z1);
change(x1,y1,z2+1);
change(x1,y2+1,z2+1);
change(x2+1,y1,z2+1);
change(x2+1,y2+1,z1);
}
else
{
scanf("%d%d%d",&x1,&y1,&z1);
printf("%d\n",sum(x1,y1,z1));
}
}
}
return 0;
}