HDU - 3584 Cube (三维树状数组 + 区间改动 + 单点求值)

Cube

Submit Status

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: 2486
Memory: 5944 KB		Time: 202 MS
Language: G++		Result: Accepted
VJ RunId: 4328542		Real RunId: 14413386
*/
//三维的与二维的一样，而二维的与一维的一样
//具体解释。本博客中解答了题目这么做的原理
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
const int MAXN = 100 + 5;
int C[MAXN][MAXN][MAXN];
int N, M;

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)) {
                C[i][j][k] ++;
            }
        }
    }
}
int query(int x, int y, int z) {
    int ret = 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)) {
                ret += C[i][j][k];
            }
        }
    }
    return ret & 1;
}
int X, x, y, z, x1, y1, z1, x2, y2, z2;
int main() {
    while(~ scanf("%d%d", &N, &M)) {
        memset(C, 0, sizeof(C));
        while(M --) {
            scanf("%d", &X);
            if(X) {
                scanf("%d%d%d%d%d%d", &x1, &y1, &z1, &x2, &y2, &z2);
                x2 ++;
                y2 ++;
                z2 ++;
                add(x1, y1, z1);
                add(x1, y2, z1);
                add(x1, y1, z2);
                add(x1, y2, z2);
                add(x2, y1, z1);
                add(x2, y2, z1);
                add(x2, y1, z2);
                add(x2, y2, z2);
            } else {
                scanf("%d%d%d", &x, &y, &z);
                printf("%d\n", query(x, y, z));
            }
        }
    }
    return 0;
}