Cube
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 2614 Accepted Submission(s): 1314
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
题意:
1.把空间上以(x1,y1,z1),(x2,y2,z2)为端点的区域所有的1变成0,0变成1.
2.求出(1,1,1)->(x,y,z)的区域的值.
转化一下思想,成为前缀和,然后&1,就知道当前位置是1还是0了.
详见代码.
1 #include <iostream> 2 #include <cstring> 3 #define N 105 4 using namespace std; 5 6 int sum[N][N][N]; 7 8 int lowbit(int x){ return x&(-x); } 9 10 void update(int a,int b,int c){ 11 for(int i = a;i<=N;i+=lowbit(i)){ 12 for(int j = b;j<=N;j+=lowbit(j)){ 13 for(int k = c; k<=N; k += lowbit(k)){ 14 sum[i][j][k]++; 15 } 16 } 17 } 18 } 19 20 int query(int a,int b,int c){ 21 int ans = 0; 22 for(int i = a;i>0;i-=lowbit(i)){ 23 for(int j = b;j>0;j-=lowbit(j)){ 24 for(int k = c;k>0;k-=lowbit(k)){ 25 ans += sum[i][j][k]; 26 } 27 } 28 } 29 return ans; 30 } 31 32 int main(){ 33 int n,m; 34 while(cin>>n>>m){ 35 memset(sum,0,sizeof(sum)); 36 int p,x,y,z,x1,y1,z1; 37 while(m--){ 38 cin>>p; 39 if(p){ 40 cin>>x>>y>>z>>x1>>y1>>z1; 41 update(x,y,z); 42 update(x,y,z1+1); 43 update(x,y1+1,z); 44 update(x1+1,y,z); 45 update(x1+1,y1+1,z); 46 update(x1+1,y,z1+1); 47 update(x,y1+1,z1+1); 48 update(x1+1,y1+1,z1+1); 49 }else{ 50 cin>>x>>y>>z; 51 cout<<(query(x,y,z)&1)<<endl; 52 } 53 } 54 } 55 return 0; 56 }