Mobile phones Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 16893 Accepted: 7789 Description Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The…
题目链接:http://poj.org/problem?id=1195 [题意] 给出一个全0的矩阵,然后一些操作 0 S:初始化矩阵,维数是S*S,值全为0,这个操作只有最开始出现一次 1 X Y A:对于矩阵的X,Y坐标增加A 2 L B R T:询问(L,B)到(R,T)区间内值的总和 3:结束对这个矩阵的操作 [思路] 二维树状数组单点更新+区域查询,可作为模板题. 注意坐标是从0开始,所以要+1 [代码] #include<cstdio> #include<cstrin…
Mobile phones Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 14489 Accepted: 6735 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 Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The squares form an S * S matrix with the rows and columns numbered from 0 to S-1. Each square contain…
Description Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The squares form an S * S matrix with the rows and columns numbered from 0 to S-1. Each square contain…
Description Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The squares form an S * S matrix with the rows and columns numbered from 0 to S-1. Each square contain…
Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The squares form an S * S matrix with the rows and columns numbered from 0 to S-1. Each square contains a base sta…
Mobile phones Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 17176 Accepted: 7920 Description Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The…
题目链接 裸二维树状数组 #include <bits/stdc++.h> const int N = 305; struct BIT_2D { int c[105][N][N], n, m; void init(int n, int m) { memset (c, 0, sizeof (c)); this->n = n; this->m = m; } void updata(int k, int x, int y, int z) { for (int i=x; i<=n;…
Matrix Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 22058 Accepted: 8219 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…
D. Iahub and Xors Iahub does not like background stories, so he'll tell you exactly what this problem asks you for. You are given a matrix a with n rows and n columns. Initially, all values of the matrix are zeros. Both rows and columns are 1-based…
题意:给你一个n*n的全0矩阵,每次有两个操作: C x1 y1 x2 y2:将(x1,y1)到(x2,y2)的矩阵全部值求反 Q x y:求出(x,y)位置的值 树状数组标准是求单点更新区间求和,但是我们处理一下就可以完美解决此问题.区间更新可以使用区间求和的方法,在更新的(x2,y2)记录+1,在更新的(x1-1,y1-1)-1(向前更新到最前方).单点求和就只需要与区间更新相反,向后求一个区间和.这样做的理由是:如果求和的点在某次更新范围内,我们+1但是不执行-1,否者要么都不执行,要么都…
Matrix Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 25004 Accepted: 9261 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…
Matrix Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 17224 Accepted: 6460 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:http://poj.org/problem?id=2155 中文题意: 给你一个初始全部为0的n*n矩阵,有如下操作 1.C x1 y1 x2 y2 把矩形(x1,y1,x2,y2)上的数全部取反,1->0,0->1 2.Q x y 求n*n矩阵的(x,y)位置上的数 题解: 先看简单的: 给你一个初始全为0的长度为n的序列,有如下操作 1.C x y 把序列x到y位取反1->0,0->1 2.Q x 求序列第x位的数 做法很简单,设计一个数组a[],a[1..i]的和表示…
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,…
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 <= i, j <= N). We can change the matrix in the following way. Given a rectangle whose upp…
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…
