F.A.Q
Hand In Hand
Online Acmers
Problem Archive
Realtime Judge Status
Authors Ranklist
 
     C/C++/Java Exams     
ACM Steps
Go to Job
Contest LiveCast
ICPC@China
Best Coder beta
VIP | STD Contests
    DIY | Web-DIY beta
Author ID 
Password 
 Register new ID

Another Brick in the Wall

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 471    Accepted Submission(s): 208


Problem Description
After years as a brick-layer, you've been called upon to analyze the structural integrity of various brick walls built by the Tetrad Corporation. Instead
of using regular-sized bricks, the Tetrad Corporation seems overly fond of bricks made out of strange shapes. The structural integrity of a wall can be
approximated by the fewest number of bricks that could be removed to create a gap from the top to the bottom. Can you determine that number for
various odd walls created by Tetrad?
 

Input
Input to this problem will begin with a line containing a single integer X (1 ≤ X ≤ 100) indicating the number of data sets. Each data set consists of
two components:

A single line, "M N" (1 ≤ M,N ≤ 20) where M and N indicate the height and width (in units), respectively, of a brick wall;
A series of M lines, each N alphabetic characters in length. Each character will indicate to which brick that unit of the wall belongs to. Note
that bricks will be contiguous; each unit of a brick will be adjacent (diagonals do not count as adjacent) to another unit of that brick. Multiple
bricks may use the same characters for their representation, but any bricks that use identical characters will not be adjacent to each other. All
letters will be uppercase.
 

Output
For each data set, output the fewest number of bricks to remove to create a gap that leads from some point at the top of the wall, to some point at the
bottom of the wall. Assume that bricks are in fixed locations and do not "fall" if bricks are removed from beneath them. A gap consists of contiguous
units of removed bricks; each unit of a gap must be adjacent (diagonals do not count) to another unit of the gap.
 

Sample Input
3 5 7 AABBCCD EFFGGHH IIJJKKL MNNOOPP QQRRSST 5 7 AABBCCD AFFBGGD IIJBKKD MNNOOPD QQRRSST 6 7 ABCDEAB ABCFEAB AEAABAB ACDAEEB FFGAHIJ KLMANOP
 

Sample Output
5 2 2
 

Source
 

Statistic | Submit | Discuss | Note
Hangzhou Dianzi University Online Judge 3.0
Copyright © 2005-2024 HDU ACM Team. All Rights Reserved.
Designer & Developer : Wang Rongtao LinLe GaoJie GanLu
Total 0.000000(s) query 1, Server time : 2024-06-17 18:42:50, Gzip enabled