Cube
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 2278 Accepted Submission(s): 1163
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
二维树状数组
和上题类似，推过来就行。
树状数组很方便的一点就是多维很好写，反之线段树的二维三维就很难写。
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
#define LL long long
const int N = 100;
int n,Q;
int sum[N+3][N+3][N+3];
int lowbit(int x)
{
return x&-x;
}
void add(int x,int y,int z)
{
for(int i=x;i<=n;i+=lowbit(i))
{
for(int j=y;j<=n;j+=lowbit(j))
{
for(int k=z;k<=n;k+=lowbit(k))
sum[i][j][k]++;
}
}
}
int query(int x,int y,int z)
{
int ans=0;
for(int i=x;i>=1;i-=lowbit(i))
{
for(int j=y;j>=1;j-=lowbit(j))
{
for(int k=z;k>=1;k-=lowbit(k))
ans+=sum[i][j][k];
}
}
return ans;
}
int main()
{
while(~scanf("%d %d",&n,&Q))
{
memset(sum,0,sizeof sum);
while(Q--)
{
int k;
scanf("%d",&k);
if(k)
{
int x1,x2,y1,y2,z1,z2;
scanf("%d %d %d %d %d %d",&x1,&y1,&z1,&x2,&y2,&z2);
add(x1,y1,z1);
add(x1,y2+1,z1);
add(x2+1,y1,z1);
add(x2+1,y2+1,z1);
add(x1,y1,z2+1);
add(x1,y2+1,z2+1);
add(x2+1,y1,z2+1);
add(x2+1,y2+1,z2+1);
}
else
{
int x,y,z;
scanf("%d %d %d",&x,&y,&z);
printf("%d\n",query(x,y,z)%2);
}
}
}
}
