首先考虑二维的情况
min(x,y)也就意味着确定最小后，另外一维肯定打满
然后最小那个如果是k的话就相当于用k*1次——这不就是行列覆盖吗，二分图秒之
三维呢？考虑到a*b*c<=5000也就是最小的那维不超过17
那么我们直接穷举那维用哪些，然后另外的就跟二维一样了
注意要优化常数
type node=record
po,next:longint;
end; var e:array[..] of node;
d:array[..,..] of longint;
p,cx,cy:array[..] of longint;
v:array[..] of boolean;
can,f:array[..] of boolean;
x,tt,j,k,len,ans,i,t,a,b,c,a0,b0,c0:longint; function max(a,b:longint):longint;
begin
if a>b then exit(a) else exit(b);
end; procedure add(x,y:longint);
begin
inc(len);
e[len].po:=y;
e[len].next:=p[x];
p[x]:=len;
end; procedure swap(var a,b:longint);
var c:longint;
begin
c:=a;
a:=b;
b:=c;
end; function dfs(x:longint):longint;
var i,y:longint;
begin
i:=p[x];
while i<> do
begin
y:=e[i].po;
if not v[y] then
begin
v[y]:=true;
if (cy[y]=-) or (dfs(cy[y])=) then
begin
cy[y]:=x;
cx[x]:=y;
exit();
end;
end;
i:=e[i].next;
end;
exit();
end; function cal(s:longint):longint;
var i,j:longint;
begin
cal:=s;
len:=;
for i:= to b do
begin
p[i]:=;
cx[i]:=-;
end;
for i:= to c do
cy[i]:=-;
for i:= to t do
if not f[d[i,a0]] then
add(d[i,b0],d[i,c0]);
for i:= to b do
if cx[i]=- then
begin
for j:= to c do
v[j]:=false;
cal:=cal+dfs(i);
if cal>=ans then exit;
end;
end; procedure work(x,s:longint);
var m:longint;
begin
if s>=ans then exit;
if x=a+ then
begin
m:=cal(s);
if m<ans then ans:=m;
end
else begin
f[x]:=false;
work(x+,s);
if can[x] then
begin
f[x]:=true;
work(x+,s+);
f[x]:=false;
end;
end;
end; begin
readln(tt);
while tt> do
begin
dec(tt);
readln(a,b,c);
a0:=; b0:=; c0:=;
t:=;
for i:= to a do
for j:= to b do
for k:= to c do
begin
read(x);
if x= then
begin
inc(t);
d[t,]:=i;
d[t,]:=j;
d[t,]:=k;
end;
end;
if (b<=a) and (b<=c) then
begin
swap(a0,b0);
swap(a,b);
end
else if (c<=a) and (c<=b) then
begin
swap(a0,c0);
swap(a,c);
end;
if b>c then
begin
swap(b0,c0);
swap(b,c);
end;
fillchar(can,sizeof(can),false);
for i:= to t do
can[d[i,a0]]:=true;
ans:=;
for i:= to a do
if can[i] then inc(ans);
fillchar(f,sizeof(f),false);
work(,);
writeln(ans);
end;
end.