HDU 3584 Cube
Cube
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 1279 Accepted Submission(s): 659
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
Author
alpc32
Source
Recommend
思路:三维树状数组,向后更新,向前求和
代码:
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cstdlib> #include <cmath> using namespace std; int map[110][110][110]; int n,m; int sum; int operation; int x_1,y_1,z_1,x_2,y_2,z_2; int posix,posiy,posiz; int lowbit(int x) { return x & (- x); } void update(int i, int j, int k, int x) { int tj, tk; while(i <= n) { tj = j; while(tj <= n) { tk = k; while(tk <= n) { map[i][tj][tk] += x; tk += lowbit(tk); } tj += lowbit(tj); } i += lowbit(i); } } int Sum(int i,int j,int k){ int ans = 0; while(i > 0) { int tempj = j; while(tempj > 0) { int tempk = k; while(tempk > 0) { ans += map[i][tempj][tempk]; tempk -= lowbit(tempk); } tempj -= lowbit(tempj); } i -= lowbit(i); } return ans; } int main() { while(~scanf("%d%d",&n,&m)) { memset(map,0,sizeof(map)); while(m --) { scanf("%d",&operation); if(operation == 1) { scanf("%d%d%d%d%d%d",&x_1,&y_1,&z_1,&x_2,&y_2,&z_2); update(x_2 + 1,y_2 + 1,z_2 + 1,1); update(x_2 + 1,y_2 + 1,z_1,1); update(x_2 + 1,y_1,z_2 + 1,1); update(x_1,y_2 + 1,z_2 + 1,1); update(x_2 + 1,y_1,z_1,1); update(x_1,y_2 + 1,z_1,1); update(x_1,y_1,z_2 + 1,1); update(x_1,y_1,z_1,1); } else { scanf("%d%d%d",&posix,&posiy,&posiz); sum = Sum(posix,posiy,posiz); printf("%d\n",sum & 1); } } } return 0; }