hdu3584--Cube（三维树状数组）Cube

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概述

Cube

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1437 Accepted Submission(s): 741

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

Author

alpc32

Source

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

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题意，n*n*n的矩阵，值为0 或者1 ，“1”操作改变(x1,y1,z1)到(x2,y2,z2)内的所有点，“0”操作问(x,y,z)的值，也就是统计改变的次数

同一维，二维一样，定义一个数组a，a[i][j][j]表示(i,j,k)到(n,n,n)的矩形中每个点都改编a[i][j][k]次，对于(x1,y1,z1)到(x2,y2,z2)的矩形改变1次，可以转化为a[x1][y1][z1] += 1，还要把多加的删去（如a(x1,y1,z1+1) += -1 等），最后保证仅对(x1,y1,z1)到(x2,y2,z2)改变

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int c[110][110][110] ;
int lowbit(int x)
{
    return x & -x ;
}
int add(int i,int j,int k,int n,int d)
{
    int x , y , z ;
    for(x = i ; x <= n ; x += lowbit(x))
        for(y = j ; y <= n ; y += lowbit(y))
            for(z = k ; z <= n ; z += lowbit(z))
                c[x][y][z] += d ;
}
int sum(int i,int j,int k)
{
    int a = 0 , x , y , z ;
    for(x = i ; x > 0 ; x -= lowbit(x))
        for(y = j ; y > 0 ; y -= lowbit(y))
            for(z = k ; z > 0 ; z -= lowbit(z))
                a += c[x][y][z];
    return a ;
}
int main()
{
    int k , n , m , x , y , z ;
    while(scanf("%d %d", &n, &m) != EOF)
    {
        memset(c,0,sizeof(c));
        while(m--)
        {
            scanf("%d %d %d %d", &k, &x, &y, &z);
            if(k == 0)
            {
                printf("%dn", sum(x,y,z)%2);
            }
            else
            {
                int x1 , y1 , z1 ;
                scanf("%d %d %d", &x1, &y1, &z1);
                add(x,y,z,n,1);
                add(x1+1,y,z,n,-1);
                add(x,y1+1,z,n,-1);
                add(x,y,z1+1,n,-1);
                add(x1+1,y1+1,z,n,1);
                add(x1+1,y,z1+1,n,1);
                add(x,y1+1,z1+1,n,1);
                add(x1+1,y1+1,z1+1,n,-1);
            }
        }
    }
    return 0;
}