HDU 3584 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): 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].
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.
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
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 else 65 { 66 scanf("%d%d%d",&x1,&y1,&z1); 67 printf("%d\n",sum(x1,y1,z1)); 68 } 69 } 70 } 71 return 0; 72 }
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HDU - 3584 Cube (三维树状数组 + 区间改动 + 单点求值)