Any hint? Is this task solved with dp bitmask, or something else?
z-ram
I need hint too..
if you know which rows you'll take you can easy decide which columns are best for that selection of rows.
So, I must consider all 2^20 combination of rows?
how to know the columns? Do the naive O(n)?
hmm shouldn't you find n - number of rows columns taken
in O(n - number of rows taken * 20 * 20 ) which is too slow ?
in O(n - number of rows taken * 20 * 20 ) which is too slow ?
pardon?
yes, you can try all combinations of rows and then in O( n ), ( n = 20 ) select columns
Ignore what I said I managed to confuse my self :D
I have done the task, but it fails on test case 13 and 14. I can't generate that kind of test case where it shows me ''wrong result''.
Does anybody know what are like one of these test cases and result?
Does anybody know what are like one of these test cases and result?
Can You tell your idea? I am looking at you code but it's hard to understand someone's code...
(Slab sam sa engleskim, zato admine izvinjavam se zbog ovog posta).
Ideja je izbacitivati red ili kolonu sa najmanje jedinica, tj. u takav red/kolonu zamenjujem sa nulama. Ono sortiranje se moze izbrisati i zameniti funkcijom koja trazi red/kolonu sa najmanje jedinica (pa bi izvršenje bilo i brže). Jedino ne mogu da kreiram primer analogan 13 ili 14 testu kada dobijam WR.
Ideja je izbacitivati red ili kolonu sa najmanje jedinica, tj. u takav red/kolonu zamenjujem sa nulama. Ono sortiranje se moze izbrisati i zameniti funkcijom koja trazi red/kolonu sa najmanje jedinica (pa bi izvršenje bilo i brže). Jedino ne mogu da kreiram primer analogan 13 ili 14 testu kada dobijam WR.
Vrlo Dobro ideja :)
Dali je sigruno da ce to rjesenje raditi? mozda neki red ima manje 1 od drugog ali ima najvise jednica na mjestima koje ima i neki drugi red? pa ti onda izbacis taj a u stvari je on bolje rjesenje?
Dali je sigruno da ce to rjesenje raditi? mozda neki red ima manje 1 od drugog ali ima najvise jednica na mjestima koje ima i neki drugi red? pa ti onda izbacis taj a u stvari je on bolje rjesenje?
Mislim da se ideja može doraditi. Ako neko ima svoje testove i rezultate, da li je voljan da mi pošalje?
ideja nije dobra..mislim jasno ti je sigurno da postoji test primjer na kojem je kriva jer je to obican greedy
OK. Thanks.
Can somebody tell me why my code isn't working?
program z_ram;
var
n,m,z,i,j,p,q,a,b:longint;
x:array [1..20,1..20] of char;
begin
readln(n);
for i:=1 to 20 do begin
for j:=1 to 20 do read(x[i,j]);
readln;
end;
m:=0;
for a:=1 to n-1 do begin
b:=n-a;
for i:=0 to 20-a do
for j:=0 to 20-b do begin
z:=0;
for p:=i+1 to i+a do
for q:=j+1 to j+b do
if (x[p,q] = '1') then z:=z+1;
if (z > m) then m:=z;
end;
end;
writeln(m);
end.
This code, when searching for 1s goes trough all matrix and gives correct results on 11 test cases, but get WR at 11th,12th,13th,14th test case ...
program z_ram;
var
n,m,z,i,j,p,q,a,b:longint;
x:array [1..20,1..20] of char;
begin
readln(n);
for i:=1 to 20 do begin
for j:=1 to 20 do read(x[i,j]);
readln;
end;
m:=0;
for a:=1 to n-1 do begin
b:=n-a;
for i:=0 to 20-a do
for j:=0 to 20-b do begin
z:=0;
for p:=i+1 to i+a do
for q:=j+1 to j+b do
if (x[p,q] = '1') then z:=z+1;
if (z > m) then m:=z;
end;
end;
writeln(m);
end.
This code, when searching for 1s goes trough all matrix and gives correct results on 11 test cases, but get WR at 11th,12th,13th,14th test case ...