HDU 3584 Cube(三位树状数组)
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Problem E
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)
Total Submission(s) : 21 Accepted Submission(s) : 12
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.
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
题意:区间更新,单点求和的三维树状数组的模板题
思路:就是数状数组模式2的三维形式(区间更新,单点求值),主要是选取哪些点进行updata,在起点增加了值后在哪些点消掉这些值(在区域外面相邻的影响到的点)
代码:
#include<iostream>#include<string>#include<cstdio>#include<algorithm>#include<cmath>#include<iomanip>#include<cstring>using namespace std;typedef long long ll;#define M 102int n,m;int c[M][M][M];int lowbit(int x){ return x&(-x);}void add(int x,int y,int z){ int i,j,k; for(i=x;i<=n;i+=lowbit(i)) for(j=y;j<=n;j+=lowbit(j)) for(k=z;k<=n;k+=lowbit(k)) { c[i][j][k]++; }}int sum(int x,int y,int z){ int i,j,k,res=0; for(i=x;i>0;i-=lowbit(i)) for(j=y;j>0;j-=lowbit(j)) for(k=z;k>0;k-=lowbit(k)) { res+=c[i][j][k]; } return res;}int main(){ios::sync_with_stdio(false); int i,s,x1,y1,z1,x2,y2,z2; while(cin>>n>>m) { memset(c,0,sizeof(c)); for(i=1;i<=m;i++) { cin>>s; if(s==1) { cin>>x1>>y1>>z1>>x2>>y2>>z2; add(x1,y1,z1); add(x2+1,y2+1,z2+1); add(x2+1,y1,z1); add(x1,y2+1,z1); add(x1,y1,z2+1); add(x2+1,y2+1,z1); add(x2+1,y1,z2+1); add(x1,y2+1,z2+1); } else { cin>>x1>>y1>>z1; printf("%d\n",sum(x1,y1,z1)&1); } } } return 0;}
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