Cube

最新推荐文章于 2024-10-17 20:25:15 发布

转载最新推荐文章于 2024-10-17 20:25:15 发布 · 79 阅读

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原文链接：http://www.cnblogs.com/zllwxm123/p/9343765.html

本文介绍了一个关于三维空间中进行特定区域的数值翻转操作及查询的问题，并提供了一种使用前缀和与低比特位操作来高效解决该问题的方法。

Cube

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 2614 Accepted Submission(s): 1314

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

题意:

１.把空间上以(x1，y1,z1),(x2,y2,z2)为端点的区域所有的1变成0，0变成1.

2.求出(1,1,1)->(x,y,z)的区域的值.

转化一下思想,成为前缀和,然后&1,就知道当前位置是1还是0了.

详见代码.

 1 #include <iostream>
 2 #include <cstring>
 3 #define N 105
 4 using namespace std;
 5 
 6 int sum[N][N][N];
 7 
 8 int lowbit(int x){ return x&(-x); }
 9 
10 void update(int a,int b,int c){
11     for(int i = a;i<=N;i+=lowbit(i)){
12         for(int j = b;j<=N;j+=lowbit(j)){
13             for(int k = c; k<=N; k += lowbit(k)){
14                 sum[i][j][k]++;
15             }
16         }
17     }
18 }
19 
20 int query(int a,int b,int c){
21     int ans = 0;
22     for(int i = a;i>0;i-=lowbit(i)){
23         for(int j = b;j>0;j-=lowbit(j)){
24             for(int k = c;k>0;k-=lowbit(k)){
25                 ans += sum[i][j][k];
26             }
27         }
28     }
29     return ans;
30 }
31 
32 int main(){
33     int n,m;
34     while(cin>>n>>m){
35         memset(sum,0,sizeof(sum));
36         int p,x,y,z,x1,y1,z1;
37         while(m--){
38             cin>>p;
39             if(p){
40                 cin>>x>>y>>z>>x1>>y1>>z1;
41                 update(x,y,z);
42                 update(x,y,z1+1);
43                 update(x,y1+1,z);
44                 update(x1+1,y,z);
45                 update(x1+1,y1+1,z);
46                 update(x1+1,y,z1+1);
47                 update(x,y1+1,z1+1);
48                 update(x1+1,y1+1,z1+1);
49             }else{
50                 cin>>x>>y>>z;
51                 cout<<(query(x,y,z)&1)<<endl;
52             }
53         }
54     }
55     return 0;
56 }

转载于:https://www.cnblogs.com/zllwxm123/p/9343765.html