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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
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 |
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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 |
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2017-12-20 09:02:47
simple multi source bfs question |
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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 |
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2017-12-11 12:33:18
solved in o(n*m*(number_of_ones)) complexity ; |
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2017-12-10 13:48:31
Finally accepted :) |
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2017-12-08 17:47:17
hmm... |
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2017-12-04 18:53:44
easy joy |
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2017-11-22 10:47:47
BFS :) calculate distance of black pixels by calling bfs from white pixels and get AC:) |
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2017-10-24 17:29:12 hanstan
BFS 0.00s :D :D :D |
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2017-10-19 11:45:22
Time limit is huge can be done using brute force with adjacency list. |