hdu3584三维树状数组

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Cube

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 2405    Accepted Submission(s): 1226

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 51 1 1 1  1 1 10 1 1 11 1 1 1  2 2 20 1 1 10 2 2 2

 

Sample Output

101

二维跟三维相似

直接上代码了

#include<iostream>#include<string.h>#include<stdio.h>using namespace std;int c[101][101][101];int n,m;int X;int x1,y1,z1,x2,y2,z2;int x,y,z;int lowbit(int x){    return x&(-x);}void update(int x,int y,int z,int w){    for(int i=x;i<=100;i=i+lowbit(i))    for(int j=y;j<=100;j=j+lowbit(j))    for(int k=z;k<=100;k=k+lowbit(k))    c[i][j][k]=c[i][j][k]+w;}int sum(int x,int y,int z){    int ans=0;    for(int i=x;i>0;i=i-lowbit(i))    for(int j=y;j>0;j=j-lowbit(j))    for(int k=z;k>0;k=k-lowbit(k))    ans=ans+c[i][j][k];    return ans;}int main(){    while(~scanf("%d%d",&n,&m))    {        memset(c,0,sizeof(c));        while(m--)        {            scanf("%d",&X);            if(X==1)            {                scanf("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2);                update(x1,y1,z1,1);                update(x2+1,y1,z1,-1);                update(x1,y2+1,z1,-1);                update(x1,y1,z2+1,-1);                update(x2+1,y2+1,z1,1);                update(x1,y2+1,z2+1,1);                update(x2+1,y1,z2+1,1);                update(x2+1,y2+1,z2+1,-1);            }            else            {                scanf("%d%d%d",&x1,&y1,&z1);                printf("%d\n",sum(x1,y1,z1)%2);            }        }    }    return 0;}