HDU 3584 Cube （三维数状数组）

Cube

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1166    Accepted Submission(s): 580

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

 

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int N=110;

int n,m,arr[N][N][N];

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

void update(int i,int j,int k,int val){
    while(i<=n){
        int tmpj=j;
        while(tmpj<=n){
            int tmpk=k;
            while(tmpk<=n){
                arr[i][tmpj][tmpk]+=val;
                tmpk+=lowbit(tmpk);
            }
            tmpj+=lowbit(tmpj);
        }
        i+=lowbit(i);
    }
}

int Sum(int i,int j,int k){
    int ans=0;
    while(i>0){
        int tmpj=j;
        while(tmpj>0){
            int tmpk=k;
            while(tmpk>0){
                ans+=arr[i][tmpj][tmpk];
                tmpk-=lowbit(tmpk);
            }
            tmpj-=lowbit(tmpj);
        }
        i-=lowbit(i);
    }
    return ans;
}

int main(){

    //freopen("input.txt","r",stdin);

    while(~scanf("%d%d",&n,&m)){
        memset(arr,0,sizeof(arr));
        int x1,y1,z1,x2,y2,z2;
        int op;
        while(m--){
            scanf("%d",&op);
            if(op==1){
                scanf("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2);
                update(x2+1, y2+1, z2+1, 1);
                update(x1, y2+1, z2+1, 1);
                update(x2+1, y1, z2+1, 1);
                update(x2+1, y2+1, z1, 1);
                update(x1, y1, z2+1, 1);
                update(x2+1, y1, z1, 1);
                update(x1, y2+1, z1, 1);
                update(x1, y1, z1, 1);
            }else {
                scanf("%d%d%d",&x1,&y1,&z1);
                printf("%d
",Sum(x1,y1,z1)&1);    //该点的值就是sum(x,y)
            }
        }
    }
    return 0;
}