【HDOJ 3584】 Cube(树状数组<区间更新，单点查询>)

【HDOJ 3584】 Cube(树状数组<区间更新，单点查询>)

Cube

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1833    Accepted Submission(s): 951

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

 

Recommend

zhouzeyong

 

跟单点更新区间查询类似 想通了就一样了

单点更新区间查询是通过更改该点所在所有不相交区间的值 区间查询时通过Sum(r)-Sum(l-1)查出 Sum(x)就是从右边界x的区间往前累加 把所有不想交区间累加起来就是区间[1,x]的点值和 Lowbite(int x) {return x&(-x)}函数是树状数组的精髓 网上有很多博客讲。。。弱就不献丑了

至于区间更新单点查询 理解了老久 后来发现就是设置增长点和削减点 更新区间[l,r]时 在l处加上 在r+1处减去 这样查询某点x时 从1~x把所有更新的值加起来 就是x处最终的点值 

此题是个三维树状数组 在更新的时候用容斥的方法设置增减点即可

代码如下:

#include <bits/stdc++.h>

using namespace std;

int bit[111][111][111];
int n;

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

int Sum(int x,int y,int z)
{
    int ans = 0;
    for(int i = x; i; i -= Lowbit(i))
        for(int j = y; j; j -= Lowbit(j))
            for(int k = z; k; k -= Lowbit(k))
                ans += bit[i][j][k];
    return ans;
}

void Add(int x,int y,int z,int data)
{
    for(int i = x; i <= n; i += Lowbit(i))
        for(int j = y; j <= n; j += Lowbit(j))
            for(int k = z; k <= n; k += Lowbit(k))
                bit[i][j][k] += data;
}

int main()
{
    int m,opt,x1,y1,z1,x2,y2,z2;

    while(~scanf("%d %d",&n,&m))
    {
        memset(bit,0,sizeof(bit));
        while(m--)
        {
            scanf("%d",&opt);
            if(opt)
            {
                scanf("%d %d %d %d %d %d",&x1,&y1,&z1,&x2,&y2,&z2);
                Add(x1,y1,z1,1);
                Add(x1,y1,z2+1,-1);
                Add(x1,y2+1,z1,-1);
                Add(x2+1,y1,z1,-1);
                Add(x2+1,y2+1,z1,1);
                Add(x2+1,y1,z2+1,1);
                Add(x1,y2+1,z2+1,1);
                Add(x2+1,y2+1,z2+1,-1);
            }
            else
            {
                scanf("%d %d %d",&x1,&y1,&z1);
                printf("%d\n",Sum(x1,y1,z1)&1);
            }
        }
    }

    return 0;
}