【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;
}