The Way Through The Forest

OpenJudge 1768: Kadane's Algorithm on 2-D Arrays Simple Version Given a 2-D matrix of integers, find the sub-matrix of this whose sum is maximized, and find that sum. The size of the matrix is no larger than 100 by 100. Analysis O(N^4) - Works, but naive     That's true, an naive O(N^4) version would work. We simply need to calculate (two dimensional!) prefix sums first. A quick review to what 2-D prefix sums are:     The 2-D prefix sum of two coordinates, X and Y, is the sum of all elements that have X-coordinate less than or equal to X, and Y coordinate less than or equal to Y.     It can be calculated through tabulation by removing overlapping areas.     Let P be the prefix sum array and V be the array of values, so the formula is as follows:     P[x, y] = V[x, y] - P[x-1, y] - P[x, y-1] + P[x-1, y-1].     We need to add the P[x-1, y-1] at the end since subtracting P[x-1, y] and P[x, y-1] we had overlapped the P[x-1, y-1]. Then, all we need to

Code Collection: VIJOS 1255 METHOD: O(n^3), Prefix Sum + O(n) Kadane's Algorithm (10/10) #include <algorithm> #include <iostream> #include <cstring> using namespace std; int aray[305][305], psum[305][305], comp[305]; const int SMALL = -22950001; int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); int R, C; cin >> R >> C; for(int i=1; i<=R; i++) for(int j=1; j<=C; j++) { cin >> aray[i][j]; if(aray[i][j] == 0) aray[i][j] = SMALL; psum[i][j] = aray[i][j] + psum[i-1][j]; } int best = aray[1][1]; for(int s=1; s<=R; s++) for(int e=s; e<=R; e++) { memset(comp, 0, sizeof comp); for(int i=1; i<=C; i++) { comp[i] = psum[e][i] - psum[s-1][i]; if(comp[i] < 0) comp[i] = SMALL; } int cmax = comp[1], cbest = comp[1]; for(int i=2; i<=C; i++) { cmax = max(comp[i], cmax+comp[i]); cbest = max(cbest

Problem Statement 1207: [HNOI2004]打鼹鼠 Time Limit:  10 Sec   Memory Limit:  162 MB Submit:  3975   Solved:  1893 [ Submit ][ Status ][ Discuss ] Description 鼹鼠是一种很喜欢挖洞的动物，但每过一定的时间，它还是喜欢把头探出到地面上来透透气的。根据这个特点阿Q编写了一个打鼹鼠的游戏：在一个n*n的网格中，在某些时刻鼹鼠会在某一个网格探出头来透透气。你可以控制一个机器人来打鼹鼠，如果i时刻鼹鼠在某个网格中出现，而机器人也处于同一网格的话，那么这个鼹鼠就会被机器人打死。而机器人每一时刻只能够移动一格或停留在原地不动。机器人的移动是指从当前所处的网格移向相邻的网格，即从坐标为（i,j）的网格移向(i-1, j),(i+1, j),(i,j-1),(i,j+1)四个网格，机器人不能走出整个n*n的网格。游戏开始时，你可以自由选定机器人的初始位置。现在你知道在一段时间内，鼹鼠出现的时间和地点，希望你编写一个程序使机器人在这一段时间内打死尽可能多的鼹鼠。 Input 第一行为n（n<=1000）, m（m<=10000），其中m表示在这一段时间内出现的鼹鼠的个数，接下来的m行每行有三个数据time,x,y表示有一只鼹鼠在游戏开始后time个时刻，在第x行第y个网格里出现了一只鼹鼠。Time按递增的顺序给出。注意同一时刻可能出现多只鼹鼠，但同一时刻同一地点只可能出现一只鼹鼠。 Output 仅包含一个正整数，表示被打死鼹鼠的最大数目 Sample Input 2 2 1 1 1 2 2 2 Sample Output 1 My Translation Moles appear on a 2-D grid on certain points at certain times. A robot knows when the moles will emerge and where they will emerge, and starts at the most optimal location possible. The robot can on

USACO 2018 February: Snow Boots (Part 2) Using  a Monotone Deque! Problem It's winter on the farm, and that means snow! There are  N N  tiles on the path from the farmhouse to the barn, conveniently numbered  1 … N 1 … N , and tile  i i  is covered in  f i f i  feet of snow. In his farmhouse cellar, Farmer John has  B B  pairs of boots, numbered  1 … B 1 … B . Some pairs are more heavy-duty than others, and some pairs are more agile than others. In particular, pair  i i  lets FJ step in snow at most  s i s i  feet deep, and lets FJ move at most  d i d i forward in each step. Farmer John starts off on tile  1 1  and must reach tile  N N  to wake up the cows. Tile  1 1  is sheltered by the farmhouse roof, and tile  N N is sheltered by the barn roof, so neither of these tiles has any snow. Help Farmer John determine which pairs of snow boots will allow him to make the trek. INPUT FORMAT (file snowboots.in): The first line contains two space-separated integers  N N  an