HDU - 3584
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
/*
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;
}
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HDU - 3584 Cube (三维树状数组 + 区间改动 + 单点求值)
原文：http://www.cnblogs.com/clnchanpin/p/7135784.html