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
Ac in one go
Ac In One Go. Wooow. This is the simplest question. You have to just apply brut force approach.
Really nice problem! Good thing it doesn't have tags.Last edit: 2021-07-16 13:49:27
ac one people should die in peace !!
Loved this question.... Take care, priority queue is not needed ;)
If ur having NZEC in java then try the same solution in c++ . Worked like a charm for me. :P
one good optimization is needed in multi source bfs. calculate the no of iterations and compare.
No need to use BFS. Applyting Brute Force works. AC in one go.
waah. multisource bfs is best.
AC in one go ! Multisource BFS :)