Looks pretty complicated :)
One problem I see right at the beginning:
for 1 1 1 your solution returns c alhough it should return Impossible.
The Minesweeper problem was the trickiest for me too, had 2 failed checks until
I got it right (my problem was that I considered things like 1 3 2 as
impossible, although it should be "c**").
Dne pondělí, 14. dubna 2014 15:58:49 UTC+2 Leopoldo Taravilse napsal(a):
> I also couldn't solve this problem (I calculated by brute force the small
> case just for making sure I advance to the next round but I couldn't solve
> C-large because my solution wasn't accepted for C-small).
>
>
>
>
> What I did was:
>
>
> if R = 1 and C = 1 return "c"
> else if R = 1 or C = 1 return something like "c...****" (or the transpose if
> C = 1)
> else if m = R*C-1 then c in [0][0] and * everywere else
>
>
> else if m = R*C - k with k in {2,3,5,7} return impossible
> else if m%2 = (R*C)%2 fill the last R-2 rows with * until I get m times '*'
> and then from C-1 to to 0 fill with '*' in the first two rows until I get m
> times '*'
>
>
> else if R = 2 or C = 2 return impossible
> else fill the last R-3 rows until m * are reached then fill the first 3 rows
> until I get m *
>
>
> This is my code in case anyone wants to read it and tell me what's wrong with
> it:
>
>
>
>
>
> char board[5][5];
>
>
> bool calc()
> {
> int m2 = m;
> if(r==1&&c==1)
> {
> board[0][0] = 'c';
>
>
> return true;
> }
> if(c==1)
> {
> forn(i,r)
> board[i][0] = '.';
> board[0][0] = 'c';
> int pos = r-1;
>
>
> while(m2>0)
> {
> board[pos][0] = '*';
> m2--;
> pos--;
> }
> return true;
> }
>
>
> if(r==1)
> {
> forn(i,c)
> board[0][i] = '.';
> board[0][0] = 'c';
> int pos = c-1;
> while(m2>0)
>
>
> {
> board[0][pos] = '*';
> m2--;
> pos--;
> }
> return true;
> }
> int sz = r*c;
>
>
> if(m == sz-1)
> {
> forn(i,r)
> forn(j,c)
> board[i][j] = '*';
> board[0][0] = 'c';
> return true;
>
>
> }
> if(m == sz-2 || m == sz-3 || m == sz-5 || m == sz-7)
> return false;
> if((sz-m)%2 == 0)
> {
> forn(i,r)
> forn(j,c)
>
>
> board[i][j] = '.';
> board[0][0] = 'c';
> int posx = r-1, posy = c-1;
> while(m2>0)
> {
> board[posx][posy] = '*';
>
>
> m2--;
> if(posx == 1 && m > 0)
> {
> board[posx-1][posy] = '*';
> m2--;
> }
>
>
> posy--;
> if(posy==-1)
> {
> posx--;
> posy = c-1;
> }
> }
> return true;
>
>
> }
> else
> {
> if(r==2||c==2)
> return false;
> forn(i,r)
> forn(j,c)
> board[i][j] = '.';
>
>
> board[0][0] = 'c';
> int posx = r-1, posy = c-1;
> while(m2>0)
> {
> board[posx][posy] = '*';
> m2--;
>
>
> if(posx == 2 && m2 > 1)
> {
> board[posx-1][posy] = '*';
> board[posx-2][posy] = '*';
> m2-=2;
>
>
> }
> posy--;
> if(posy==-1)
> {
> posx--;
> posy = c-1;
> }
> }
>
>
> return true;
> }
> }
>
>
>
> On Sun, Apr 13, 2014 at 8:26 AM, Leopoldo Salvo <[email protected]> wrote:
>
>
> HI,
>
>
>
> I could not get this one to pass either, i still don't understand why, it
> must be the way i fill the board (I ran the actual recursion algorigthm on
> both failures and successes and checked the number of cells left to be
> revealed and they were consistent with my results)
>
>
>
>
>
> If R or C == 1 then it is always possible
>
>
>
> otherwise, i calculate a rectangle top left of the board with width C-2 and
> height R-2, i call that a buffer zone, as long as M fits in that buffer zone
> it is possible (I fill it column by column: from top to bottom and from left
> to right)
>
>
>
>
>
> If M > bufferzone then i start filling the rectangle underneath it, so at max
> 2 rows height, and at most C-2 columns width. In this area i say the board
> will be ok as long as the number of mines in this container is even. I fill
> this rectangle from top to bottom and from left to right.
>
>
>
>
>
> after this lower rectangle is filled, if there are more mines remaining i do
> the same thing with the rectangle right of the buffer zone (this is a
> rectangle with at most 2 columns and R-2 rows). I fill this from top to
> bottom, from left to right. As long as the number of mines in this rectangle
> is even, it is ok.
>
>
>
>
>
> at this point there could be at most 3 more mines (requirements say M < R*C.
>
>
>
> IF remaining = 3 then it is possible (all mines except the clicked c),
> otherwise it is impossible.
>
>
>
> I am still baffled why it didn't pass, there must be some corner case or the
> way i fill it is not correct, any ideas?
>
>
>
> I suppose ill have to just run the data set against one of the passed ones
> and compare
>
>
>
>
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