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

HDU - 3584

Cube

Time Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u

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 51 1 1 1  1 1 10 1 1 11 1 1 1  2 2 20 1 1 10 2 2 2 

 

Sample Output

101 

 

/*Author: 2486Memory: 5944 KBTime: 202 MSLanguage: G++Result: AcceptedVJ RunId: 4328542Real RunId: 14413386*///三维的与二维的一样，而二维的与一维的一样//详解，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;}