HDU 3584 Cube （三维树状数组）

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

三维树状数组。

。

加一个for循环就ok

#include <iostream>
#include <cstring>
#include <cstdio>
//#include <cmath>
#include <set>
#include <stack>
#include <cctype>
#include <algorithm>
#define lson o<<1, l, m
#define rson o<<1|1, m+1, r
using namespace std;
typedef long long LL;
const int mod = 99999997;
const int MAX = 1000000000;
const int maxn = 1005;
int n, q, x1, y1, z1, x2, y2, z2, op;
int c[101][101][101];
void add(int x, int y, int z) {for(int i = x; i <= n; i += i&-i)for(int j = y; j <= n; j += j&-j)for(int k = z; k <= n; k += k&-k)c[i][j][k]++;
}
int query(int x, int y, int z) {int sum = 0;for(int i = x; i > 0; i -= i&-i)for(int j = y; j > 0; j -= j&-j)for(int k = z; k > 0; k -= k&-k)sum += c[i][j][k];return sum;
}
int main()
{//freopen("in.txt", "r", stdin);while(cin >> n >> q) {memset(c, 0, sizeof(c));while(q--) {scanf("%d%d%d%d", &op, &x1, &y1, &z1);if(op) {scanf("%d%d%d", &x2, &y2, &z2);x2++, y2++, z2++;add(x1, y1, z1);add(x1, y1, z2);add(x1, y2, z1);add(x2, y1, z1);add(x1, y2, z2);add(x2, y1, z2);add(x2, y2, z1);add(x2, y2, z2);} else printf("%d\n", query(x1, y1, z1) & 1);}}return 0;
}

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