fashion_show
``` ```
% 2017 Qualification Round, Problem D, in Picat, by Afa Zhou
% Fashion Show

import mip, util.

main =>
OS = open("out",write),
foreach (TC in 1..T)
[N,M] = [to_int(Token) : Token in read_line().split()],
once dc(TC,N,M,OS)
end,
close(OS).

dc(TC,N,M,OS) =>
Mx0 = new_array(N,N),
Mp0 = new_array(N,N),
Mx = new_array(N,N), Mx :: 0..1,
Mp = new_array(N,N), Mp :: 0..1,
foreach (_I in 1..M)
MToken = [Model],
R = RToken.to_int(),
C = CToken.to_int(),
fill(Model,R,C,Mx0,Mp0,Mx,Mp)
end,
bind_vars(Mx0,0), bind_vars(Mp0,0),  % a cell is 0 if it is not occupied in the given configuration
foreach (I in 1..N)
sum([Mx[I,C] : C in 1..N]) #= 1,
sum([Mx[R,I] : R in 1..N]) #= 1
end,
foreach(K in 1-N..N-1)
sum([Mp[R,C] : R in 1..N, C = R-K, C >= 1, C =< N]) #=< 1
end,
foreach(K in 2..2*N)
sum([Mp[R,C] : R in 1..N, C = K-R, C >= 1, C =< N]) #=< 1
end,
sum([Mp[R,C] : R in 1..N, C in 1..N]) #= NumP,
solve(\$[max(NumP)],(Mp,Mx)),
output(TC,NumP,N,Mx0,Mp0,Mx,Mp,OS).

fill('o',R,C,Mx0,Mp0,Mx,Mp) =>
Mx0[R,C] = 1, Mp0[R,C] = 1, Mx[R,C] = 1, Mp[R,C] = 1.
fill('+',R,C,_Mx0,Mp0,_Mx,Mp) =>
Mp0[R,C] = 1, Mp[R,C] = 1.
fill('x',R,C,Mx0,_Mp0,Mx,_Mp) =>
Mx0[R,C] = 1, Mx[R,C] = 1.

output(TC,Nump,N,Mx0,Mp0,Mx,Mp,OS) =>
Count = 0,
foreach (R in 1..N, C in 1..N)
if (Mx[R,C] != Mx0[R,C] || Mp[R,C] != Mp0[R,C]) then
Count := Count+1
end
end,
printf(OS,"Case #%w: %w %w\n", TC,Nump+N,Count),
foreach (R in 1..N, C in 1..N)
if (Mx[R,C] != Mx0[R,C] || Mp[R,C] != Mp0[R,C]) then
if (Mx[R,C] == 1 && Mp[R,C] == 1) then
Model = 'o'
elseif (Mx[R,C] == 1) then
Model = 'x'
else
Model = '+'
end,
printf(OS,"%w %w %w\n",Model,R,C)
end
end.
``````