Find the best immediate move in a "match-3" game

Question

Your challenge today is to take input like this:

fbcfbee
ffcabbe
debceec
bccabbe
edcfbcd
daeaafc
eebcbeb

And output the best possible move in a Bejeweled-like game that will match three or more letters, like this (note the capital B and C):

fbcfbee
ffcabbe
deBCeec
bccabbe
edcfbcd
daeaafc
eebcbeb

Full specifications:

• The input will be n lines of n lowercase letters each (where n could be any number).
• The output will be the best move you could make in a match-3 game, with the two letters you want to swap capitalized.

• Matches should have the following priority (in these examples, . indicates a square that doesn't matter):

  1. Five-in-a-row

     xxYxx
     ..X..
     

  2. Broken five-in-a-row

     X..
     Yxx
     x..
     x..
     

     or

     .X.
     xYx
     .x.
     .x.
     

  3. Four-in-a-row

     xYxx
     .X..
     

  4. Three-in-a-row

     xYx
     .X.
     

  You must find the match of the highest priority and output it.

• If there are multiple matches of the same priority, you can output any one of them.
• There will always be at least one match (your program can break if there are no matches, or do anything you want).
• I/O can be in any reasonable format (stdin/out, reading and writing files, function arguments/return values, dialog boxes, etc.) but NOT hardcoded (like x="[insert input here]").
• This is code-golf so shortest code in bytes wins. If you use any network access for some reason, all bytes downloaded from the network count against your score.

Answer 1

Python3.4, 772

(Using tabs for indentation, instead of spaces.)

import sys,itertools as I
B=[]
for l in sys.stdin:
    l=l.rstrip()
    B.append(list(l))
Z=len(B[0])
F=T=None
R=range
N=min
X=max
P=I.product
S=0
def C(I,J,K,L):
    global F,T,S
    if K<0 or K>=Z or L<0 or L>=Z: return
    B[I][J],B[K][L]=B[K][L],B[I][J]
    h=v=1
    m=B[K][L]
    for i in R(K+1,N(Z,K+5)):
        if B[i][L]!=m:break
        v+=1
    for i in R(K-1,X(0,K-5),-1):
        if B[i][L]!=m:break
        v+=1
    for j in R(L+1,N(Z,L+5)):
        if B[K][j]!=m:break
        h+=1
    for j in R(L-1,X(0,L-5),-1):
        if B[K][j]!=m:break
        h+=1
    c=X(h,v)*2
    if N(h,v)>=3:c+=N(h,v)
    if c>S:S=c;F=I,J;T=K,L
    B[I][J],B[K][L]=B[K][L],B[I][J]
for i,j in P(reversed(R(Z)),R(Z)):
    for d,e in (1,0),(0,-1),(0,1),(-1,0):
        C(i,j,i+d,j+e)
for i,j in P(R(Z),R(Z)):
    c=B[i][j]
    if (i,j)in(F,T):c=c.upper()
    print(c,end=('',"\n")[j==Z-1])