BITMAP - Bitmap

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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:

d(p1,p2)=|i1-i2|+|j1-j2|.

Task

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.

Input

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.

Output

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.

Example

Sample input:
1
3 4
0001
0011
0110

Sample output:
3 2 1 0
2 1 0 0
1 0 0 1

hide comments
grylls: 2018-03-06 18:50:27

AC through DP but i was thinking of multi source BFS Solution. if anyone submitted Plz send the code at gmail

ankitsaininit@gmail.com

hai_dee: 2018-02-22 10:53:28

Great question, but PLEASE change the wording regarding the input format. It cost me 3 RE's :(

nitishk24: 2018-01-05 02:11:33

DP solution: Break in 2 parts, one traversal can pass traverses bottom right while another up left
ans is minimum of 2 dp

frozen7: 2017-12-20 09:02:47

simple multi source bfs question

karthik1997: 2017-12-17 17:39:51

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
sinersnvrsleep: 2017-12-11 12:33:18

solved in o(n*m*(number_of_ones)) complexity ;

ehsanhy: 2017-12-10 13:48:31

Finally accepted :)

malish_228: 2017-12-08 17:47:17

hmm...

malish_228: 2017-12-04 18:53:44

easy joy

anurag_tangri: 2017-11-22 10:47:47

BFS :) calculate distance of black pixels by calling bfs from white pixels and get AC:)


Added by:Piotr Ɓowiec
Date:2004-09-13
Time limit:4s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All
Resource:6th Polish Olympiad in Informatics, stage 2