BITMAP - Bitmap
There is given a rectangular bitmap of size n*m. Each pixel of the bitmap is either white or black, but at least one is white. The pixel in i-th line and j-th column is called the pixel (i,j). The distance between two pixels p1=(i1,j1) and p2=(i2,j2) is defined as:
Write a program which:
- reads the description of the bitmap from the standard input,
- for each pixel, computes the distance to the nearest white pixel,
- writes the results to the standard output.
The number of test cases t is in the first line of input, then t test cases follow separated by an empty line. In the first line of each test case there is a pair of integer numbers n, m separated by a single space, 1<=n <=182, 1<=m<=182. In each of the following n lines of the test case exactly one zero-one word of length m, the description of one line of the bitmap, is written. On the j-th position in the line (i+1), 1 <= i <= n, 1 <= j <= m, is '1' if, and only if the pixel (i,j) is white.
In the i-th line for each test case, 1<=i<=n, there should be written m integers f(i,1),...,f(i,m) separated by single spaces, where f(i,j) is the distance from the pixel (i,j) to the nearest white pixel.
Sample input: 1 3 4 0001 0011 0110 Sample output: 3 2 1 0 2 1 0 0 1 0 0 1
hint - use multisource BFS!! :-)
ac in one go
I am getting WA while using brute force all TC worked for me Help!
Awesome problem. I remember looking at it a year ago, unable to figure out a decent approach; tonight I immediately came up with 2 different ones, invented third in the process and have a vague idea how to do it better still. AC with each -- love problems that allow such experiments. No points for this, but a lot of fun for sure!
AC through DP but i was thinking of multi source BFS Solution. if anyone submitted Plz send the code at gmail
Great question, but PLEASE change the wording regarding the input format. It cost me 3 RE's :(
DP solution: Break in 2 parts, one traversal can pass traverses bottom right while another up left
simple multi source bfs question
Nice Problem Try using 2pass DP , (top down and bottom up passes ) and its complexity is O(M*N) . Would never go TLE , if m and n are increased by 100 folds => to 10000 , while bfs surely results in TLE.Last edit: 2017-12-17 17:41:34