HDU - 3584 Cube Time Limit: 1000MS   Memory Limit: 65536KB   64bit IO Format: %I64d & %I64u Submit Status 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. I…

题意:给一个三维数组n*n*n,初始都为0,每次有两个操作: 1. 翻转(x1,y1,z1) -> (x2,y2,z2) 0. 查询A[x][y][z] (A为该数组) 解法:树状数组维护操作次数,一个数被操作偶数次则相当于没被操作. 每次更新时在8个位置更新: .相当于8个二进制数:000,001,010,011,100,101,110,111. (我是由二维推过来的) 其实不用有的为-1,直接1也行,因为反正会改变奇偶性. 代码: #include <iostream> #inclu…

Minimum Inversion Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 18543    Accepted Submission(s): 11246 Problem Description The inversion number of a given number sequence a1, a2, ..., a…

Cube Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 1166    Accepted Submission(s): 580 Problem Description Given an N*N*N cube A, whose elements are either 0 or 1. A[i, j, k] means the number…

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"…

题意:还是那篇论文里面讲到的,三维树状数组http://wenku.baidu.com/view/1e51750abb68a98271fefaa8画个立方体出来对照一下好想一点 #include<iostream> #include<cstdio> #include<cstring> #include <cmath> #include<stack> #include<vector> #include<map> #inclu…

三维树状数组模版.优化不动了. #include <set> #include <map> #include <stack> #include <cmath> #include <queue> #include <cstdio> #include <string> #include <vector> #include <iomanip> #include <cstring> #inclu…

题意:... 析:可以直接用数状数组进行模拟,也可以用线段树. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #inc…

1470 最简单的三维树状数组 #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<stdlib.h> using namespace std; #define lowbit(x) x&(-x) ][][],n; int getsum(int x,int y,int z) { ,i,j,g; ; i -= lowbit(i)…

题意:对于两个区间,[si,ei] 和 [sj,ej],若 si <= sj and ei >= ej and ei - si > ej - sj 则说明区间 [si,ei] 比 [sj,ej] 强.对于每个区间,求出比它强的区间的个数. 解题思路:先将每个区间按 e 降序排列,在按 s 升序排列.则对于每个区间而言,比它强的区间的区间一定位于它的前面. 利用数状数组求每个区间[si,ei]前面 满足条件的区间[sj,ej]个数(条件:ej<=ei),再减去前面的和它相同的区间的个…