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HDU 3584 Cube（三位树状数组）

2024-08-14 07:06:51   220人阅读

Cube

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1949    Accepted Submission(s): 1013


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
 

 

Author
alpc32
 

 

Source
2010 ACM-ICPC Multi-University Training Contest（15）——Host by NUDT
 

 

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/*题意实在太模糊了，两种操作0：单点查询，这个点的数字是多少1：将某一区间内的数字都反转；只需要维护反转次数的前缀和就行了编译器出毛病了，宏定义不行，调用函数也不行*/#include<iostream>#include<stdio.h>#include<string.h>#define N 110//#define lowbit(x) x&(-x)using namespace std;int n,m;int c[N][N][N];int lowbit(int x){    return x&(-x);}void update(int x,int y,int z,int val){    while(x<=n)    {        int j=y;        while(j<=n)        {            int k=z;            while(k<=n)            {                c[x][j][k]+=val;                k+=lowbit(k);            }            j+=lowbit(j);        }        x+=lowbit(x);    }}int getsum(int x,int y,int z){    int s=0;    while(x>0)    {        int j=y;        while(j>0)        {            int k=z;            while(k>0)            {                s+=c[x][j][k];                k-=lowbit(k);            }            j-=lowbit(j);        }        x-=lowbit(x);    }    return s;}int main(){    //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);    while(scanf("%d%d",&n,&m)!=EOF)    {        memset(c,0,sizeof c);        int op,x1,y1,z1,x2,y2,z2;        while(m--)        {            scanf("%d",&op);            if(op)            {                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);                //cout<<"getsum(x1,y1,z1)="<<getsum(x1,y1,z1)<<endl;                printf("%d\n",getsum(x1,y1,z1)&1);            }        }    }    return 0;}

 

HDU 3584 Cube（三位树状数组）


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