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]...

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 opti...

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 t...