树状数组 / 二维树状数组

https://www.bnuoj.com/v3/contest_show.php?cid=9148#problem/F [题意] 给定一个矩阵,每个格子的初始值为1.现在可以对矩阵有四种操作: A x y n1 :给格点(x,y)的值加n1 D x y n1: 给格点(x,y)的值减n1,如果现在格点的值不够n1,把格点置0 M x1 y1 x2 y2:(x1,y1)移动给(x2,y2)n1个 S x1 y1 x2 y2 查询子矩阵的和 [思路] 当然是二维树状数组 但是一定要注意:lowbi

题目链接:POJ 1195 题意: 给出一个S*S的矩阵(行.列号从1开始),每个元素初始值为0,有两种操作:一种是第X行第Y列元素值加A:另一种是查询给定范围矩阵的所有元素之和(L<=X<=R,B<=Y<=T). 分析: 查询给定范围矩阵的所有元素之和是二维区间和,可以转换为二维前缀和求值.类比一维前缀和求法,二维区间和S(L, B, R, T) = S(1, 1, R, T) - S(1 ,1, L-1, T) - S(1, 1, R, B-1) + S(1, 1, L-1,

Matrix Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 20599   Accepted: 7673 Description Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (1

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5465 Clarke and puzzle Accepts: 42 Submissions: 269 Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) 问题描述 克拉克是一名人格分裂患者.某一天,有两个克拉克(aa和bb)在玩一个方格游戏. 这个方格是一个n*mn∗m的矩阵,每个格子里有一

将坐标旋转45度就可以得到正方形,可以用二维树状数组求解... 为了节省内存,提前将树状数组中会被更新的点全都存下来,并离散化 Crowd Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1199    Accepted Submission(s): 282 Problem Description City F in the south

Description Seiji Hayashi had been a professor of the Nisshinkan Samurai School in the domain of Aizu for a long time in the 18th century. In order to reward him for his meritorious career in education, Katanobu Matsudaira, the lord of the domain of

Mobile phones Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 14288   Accepted: 6642 Description Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The

Description Every year, Farmer John's N (1 <= N <= 20,000) cows attend "MooFest",a social gathering of cows from around the world. MooFest involves a variety of events including haybale stacking, fence jumping, pin the tail on the farmer,

http://acm.hdu.edu.cn/showproblem.php?pid=4456 第一道二维树状数组就这么麻烦,题目要计算的是一个菱形范围内的和,于是可以把原来的坐标系旋转45度,就是求一个正方形范围内的和,这里还涉及到坐标系平移和放大,由于题目数据较大,用了离散化来保存需要处理的点,放在h数组内,对应的二维树状数组存在tree数组内. #include<iostream> #include<cstring> #include<cstdio> #includ