三维三维三维树状数组模板题 hdu 3584 Cube

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Cube

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1757    Accepted Submission(s): 917

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

2010 ACM-ICPC Multi-University Training Contest（15）——Host by NUDT

 

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思路: 三维树状数组

分析:

1 点击打开查看论文  建议先看看这篇论文，然后就懂了，裸的三维树状数组

代码:

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

typedef long long int64;
const int MAXN = 1e2+10;

int n , m;
int64 treeNum[MAXN][MAXN][MAXN];

int lowbit(int x){
    return x&(-x);
}

int64 getSum(int x , int y , int z){
    int64 sum = 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))
                sum += treeNum[i][j][k];
    return sum; 
}

void add(int x , int y , int z , int val){
    for(int i = x ; i < MAXN ; i += lowbit(i))
        for(int j = y ; j < MAXN ; j += lowbit(j))
            for(int k = z ; k < MAXN ; k += lowbit(k))
                treeNum[i][j][k] += val;
}

void solve(){
    memset(treeNum , 0 , sizeof(treeNum));
    int mark;
    int x1 , y1 , z1;
    int x2 , y2 , z2;
    while(m--){
         scanf("%d" , &mark);
         if(mark == 1){
             scanf("%d%d%d" , &x1 , &y1 , &z1);
             scanf("%d%d%d" , &x2 , &y2 , &z2);
             x1++ , y1++ , z1++;
             x2++ , y2++ , z2++;
             // up
             add(x1 , y1 , z1 , 1); 
             add(x1 , y2+1 , z1 , -1); 
             add(x2+1 , y1 , z1 , -1); 
             add(x2+1 , y2+1 , z1 , 1); 
             // down
             add(x1 , y1 , z2+1 , -1); 
             add(x1 , y2+1 , z2+1 , 1); 
             add(x2+1 , y1 , z2+1 , 1); 
             add(x2+1 , y2+1 , z2+1 , -1); 
         }
         else{
             scanf("%d%d%d" , &x1 , &y1 , &z1);
             x1++ , y1++ , z1++;
             int ans = getSum(x1 , y1 , z1)%2;
             printf("%d\n" , ans);
         }
    }
}

int main(){
    while(scanf("%d%d" , &n , &m) != EOF)
         solve(); 
    return 0;
}