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 shot do not use DFS algo if you want to escape TLE just bfs traversal is quite enough for this problem
when i solve using bfs,it gets tle
learned multisource bfs
no need to confuse with multisource bfs just store all the indices of '1' in queue and then apply BFS
AC in 1 go :D
Do a Multi-source bfs from white pixels, it works like a charm!
why tle for bfs??
DO NOT LISTEN TO OTHERS WHILE SOLVING PROBLEM....everyone was saying multisource bfs and i feared cause i have not even heard of that .............i applied simple bfs approach and it got accepted...so ddont listen to anyone else and keep solving your problem....hAPPY Coding!!
Use multisource bfs :)
input can be: