LABYR1  Labyrinth
The northern part of the Pyramid contains a very large and complicated labyrinth. The labyrinth is divided into square blocks, each of them either filled by rock, or free. There is also a little hook on the floor in the center of every free block. The ACM have found that two of the hooks must be connected by a rope that runs through the hooks in every block on the path between the connected ones. When the rope is fastened, a secret door opens. The problem is that we do not know which hooks to connect. That means also that the neccessary length of the rope is unknown. Your task is to determine the maximum length of the rope we could need for a given labyrinth.
Input
The input consists of T test cases. The number of them (T) is given on
the first line of the input file.
Each test case begins with a line containing two integers C and R
(3 <= C,R <= 1000) indicating the number of columns and rows. Then
exactly R lines follow, each containing C characters. These characters
specify the labyrinth. Each of them is either a hash mark (#
) or a period
(.
). Hash marks represent rocks, periods are free blocks. It is
possible to walk between neighbouring blocks only, where neighbouring
blocks are blocks sharing a common side. We cannot walk
diagonally and we cannot step out of the labyrinth.
The labyrinth is designed in such a way that there is exactly one path between any two free blocks. Consequently, if we find the proper hooks to connect, it is easy to find the right path connecting them.
Output
Your program must print exactly one line of output for each test case. The
line must contain the sentence
"Maximum rope length is X.
" where Xis the
length of the longest path between any two free blocks, measured in blocks.
Example
Sample Input: 2 3 3 ### #.# ### 7 6 ####### #.#.### #.#.### #.#.#.# #.....# ####### Sample output: Maximum rope length is 0. Maximum rope length is 8.Warning: large Input/Output data, be careful with certain languages
Added by:  Adrian Kosowski 
Date:  20040606 
Time limit:  5s 
Source limit:  50000B 
Memory limit:  1536MB 
Cluster:  Cube (Intel Pentium G860 3GHz) 
Languages:  All except: SCM chicken 
Resource:  ACM Central European Programming Contest, Prague 1999 
hide comments
tarunsai:
20150706 09:21:23
very easy BFS+BFS :) 

SangKuan:
20150701 05:30:55
17 wa,now got ac...o(∩∩)o...use dfs or bfs,not esay 

sujit yadav:
20150621 00:55:37
i hate c++14 compiler gives me 4 compilation error .. wtf y? :( 

peeyush taneja:
20150616 16:40:57
After 3 wa finally AC 

Dipti Singhal:
20150612 11:58:42
Giving Runtime Error!


Rishabh:
20150608 11:50:46
"there is exactly one path between any two free blocks". Does this means that there will be only one connected component (island) ? 

Shubham Bansal:
20150518 16:04:01
sometimes something easy may also be misleading :P


eightnoteight:
20150109 07:38:32
after solving this question, try to solve PT07Z


mayank:
20141115 15:06:30
This test case might be of help.


Mitch Schwartz:
20140808 15:34:53
@arbit: Read the problem statement carefully. 