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HDU 3584 树状数组

2024-08-14 23:22:47   224人阅读

Cube

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1956    Accepted Submission(s): 1017


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
题意：三维坐标内，每一个点初始值为0，每次改变1->0,0->1,1 111 222，表示改变（1,1,1）~（2，2,2）范围的点，0 111 表示（1,1,1）点的值几。
代码：
 1 /* 2 这题挺巧妙，树状数组求和，因为变1次和变三次一样所以最后变得次数是奇数结果就是1，偶数结果就是0； 3 变数的时候注意是三维的，三个坐标要全部在(x1,y1,z1)~(x2,y2,z2)内。 4 */ 5 #include<iostream> 6 #include<cstdio> 7 #include<cstring> 8 using namespace std; 9 int A[102][102][102];10 int n,m;11 int lowbit(int a)12 {13     return a&(-a);14 }15 void change(int a1,int b1,int c1)16 {17     for(int i=a1;i<=n;i+=lowbit(i))18     {19         for(int j=b1;j<=n;j+=lowbit(j))20         {21             for(int k=c1;k<=n;k+=lowbit(k))22             {23                 A[i][j][k]++;24             }25         }26     }27 }28 int sum(int a1,int b1,int c1)29 {30     int ans=0;31     for(int i=a1;i>0;i-=lowbit(i))32     {33         for(int j=b1;j>0;j-=lowbit(j))34         {35             for(int k=c1;k>0;k-=lowbit(k))36             {37                 ans+=A[i][j][k];38             }39         }40     }41     return ans&1;42 }43 int main()44 {45     int x,x1,y1,z1,x2,y2,z2;46     while(scanf("%d%d",&n,&m)!=EOF)47     {48         memset(A,0,sizeof(A));49         while(m--)50         {51             scanf("%d",&x);52             if(x)53             {54                 scanf("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2);55                 change(x1,y1,z1);56                 change(x2+1,y2+1,z2+1);57                 change(x2+1,y1,z1);58                 change(x1,y2+1,z1);59                 change(x1,y1,z2+1);60                 change(x1,y2+1,z2+1);61                 change(x2+1,y1,z2+1);62                 change(x2+1,y2+1,z1);            63             }64             else65             {66                 scanf("%d%d%d",&x1,&y1,&z1);67                 printf("%d\n",sum(x1,y1,z1));68             }69         }70     }71     return 0;72 }

 

 

HDU 3584 树状数组


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