HDU 3584 Cube（三维树状数组） Cube

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我是靠谱客的博主甜美电灯胆，最近开发中收集的这篇文章主要介绍HDU 3584 Cube（三维树状数组） Cube，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

Cube

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1660 Accepted Submission(s): 865

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

Sample Output


1
0
1

Author

alpc32

Source

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

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题意： N*N*N的立方体，元素为0或1（初始为0）。1操作将（x1,y1,z1）到（x2,y2,z2）之间的元素取反；0操作是询问（x,y,z）为0或1。

题解：三维树状数组，原理跟一维，二维一样，可以参考http://blog.csdn.net/acm_baihuzi/article/details/46819049

#include<cstring>
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstdio>
#define N 110
using namespace std;
int n,q;
int bit[N][N][N];
int getsum(int a,int b,int c) {
int sum=0;
for(int i=a; i>0; i-=i&-i) {
for(int j=b; j>0; j-=j&-j) {
for(int k=c; k>0; k-=k&-k) {
sum+=bit[i][j][k];
}
}
}
return sum;
}
void add(int a,int b,int c) {
for(int i=a; i<=n; i+=i&-i) {
for(int j=b; j<=n; j+=j&-j) {
for(int k=c; k<=n; k+=k&-k) {
bit[i][j][k]^=1;
}
}
}
}
int main() {
//freopen("test.in","r",stdin);
while(cin>>n>>q) {
memset(bit,0,sizeof bit);
int op,x2,y2,z2,x1,y1,z1;
while(q--) {
scanf("%d%d%d%d",&op,&x1,&y1,&z1);
if(op==0) {
printf("%dn",getsum(x1,y1,z1)%2);
continue;
}
scanf("%d%d%d",&x2,&y2,&z2);
add(x1,y1,z1);
add(x1,y1,z2+1);
add(x1,y2+1,z1);
add(x1,y2+1,z2+1);
add(x2+1,y1,z1);
add(x2+1,y1,z2+1);
add(x2+1,y2+1,z1);
add(x2+1,y2+1,z2+1);
}
}
return 0;
}