HDU 3584 Cube（三位树状数组） Problem E

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概述

Problem E

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 21 Accepted Submission(s) : 12

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：区间更新，单点求和的三维树状数组的模板题

思路:就是数状数组模式2的三维形式（区间更新，单点求值），主要是选取哪些点进行updata，在起点增加了值后在哪些点消掉这些值（在区域外面相邻的影响到的点）

代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<iomanip>
#include<cstring>
using namespace std;
typedef long long ll;
#define M 102
int n,m;
int c[M][M][M];
int lowbit(int x)
{
    return x&(-x);
}
void add(int x,int y,int z)
{
    int i,j,k;
    for(i=x;i<=n;i+=lowbit(i))
        for(j=y;j<=n;j+=lowbit(j))
            for(k=z;k<=n;k+=lowbit(k))
    {
        c[i][j][k]++;
    }
}
int sum(int x,int y,int z)
{
    int i,j,k,res=0;
    for(i=x;i>0;i-=lowbit(i))
        for(j=y;j>0;j-=lowbit(j))
            for(k=z;k>0;k-=lowbit(k))
    {
        res+=c[i][j][k];
    }
    return res;
}
int main()
{
	ios::sync_with_stdio(false);
    int i,s,x1,y1,z1,x2,y2,z2;
    while(cin>>n>>m)
    {
        memset(c,0,sizeof(c));
        for(i=1;i<=m;i++)
        {
            cin>>s;
            if(s==1)
            {
                cin>>x1>>y1>>z1>>x2>>y2>>z2;
                add(x1,y1,z1);
                add(x2+1,y2+1,z2+1);
                add(x2+1,y1,z1);
                add(x1,y2+1,z1);
                add(x1,y1,z2+1);
                add(x2+1,y2+1,z1);
                add(x2+1,y1,z2+1);
                add(x1,y2+1,z2+1);
            }
            else
            {
                cin>>x1>>y1>>z1;
                printf("%dn",sum(x1,y1,z1)&1);
            }
        }
    }
    return 0;
}