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:
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
danicoder_7:
2023-11-03 08:47:38
I Have done correct bfs with memoization but it is still giving TLE whyyyy
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ihuntuuu:
2022-01-27 08:02:42
Multisource BFS ! |
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fahd_hakem:
2021-10-31 21:37:29
Ac in one go |
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musky:
2021-08-06 06:26:54
Ac In One Go. Wooow. This is the simplest question. You have to just apply brut force approach. |
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yasser1110:
2021-07-16 13:49:00
Really nice problem! Good thing it doesn't have tags. Last edit: 2021-07-16 13:49:27 |
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rec28:
2021-06-15 17:58:23
ac one people should die in peace !! |
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wslord:
2020-12-29 21:05:32
Loved this question.... Take care, priority queue is not needed ;) |
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kanishkverma_1:
2020-11-17 17:29:49
If ur having NZEC in java then try the same solution in c++ . Worked like a charm for me. :P |
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varuntumbe:
2020-10-23 10:01:08
one good optimization is needed in multi source bfs. calculate the no of iterations and compare. |
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gnomegeek:
2020-08-04 23:42:49
No need to use BFS. Applyting Brute Force works. AC in one go. |
Added by: | Piotr Ćowiec |
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 |