10977  Enchanted Forest
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10977  Enchanted Forest
I tried to solve this problem using bfs and i had lot of WA
Can anyone give me some tricky case .
Please Help.
Can anyone give me some tricky case .
Please Help.
somebody check my code.. please.. ;;
i do BFS..
and don't know why WA..
if position (i, j) is blocked, i checked map[j] = 1
i do BFS..
and don't know why WA..
if position (i, j) is blocked, i checked map[j] = 1
Code: Select all
CUT AFTER AC
Last edited by helloneo on Thu Dec 29, 2005 12:23 pm, edited 2 times in total.

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 Posts: 1
 Joined: Thu Dec 29, 2005 8:47 am
 Location: ICU, Korea
For each case, if some "dangerous" places are unavoidable, print "Impossible."Code: Select all
printf("Impossible\n");
Yes, that was the only mistake in that code.
Now, I think my BFS is exactly the same, but it keeps timing out (in Java):
Sq is just Point with distance from (0,0).
Note that the queue above is actually Vector (has some sync overhead, but is still not that bad). We can't use any other builtin types on OJ (except Hastable), so it is kind of fun.
From what I gathered, it takes 0.2 secs per grid! (Not at my home PC, those 19900 path grids take the longest  0.01sec, otherwise, it can do ~1000 empty 200x200 in sec)
I tried inserting points that are closer to the end in front of the queue(sort of A*), but that didn't change it much.
I used exactly the same BFS algorithm to fill in 10267, I didn't run out of time there.
Can I improve it somehow? Or is it just OJ Java thing?
Darko
Now, I think my BFS is exactly the same, but it keeps timing out (in Java):
Code: Select all
private int bfs() {
queue.addElement(new Sq(0, 0, 0));
grid[0][0] = 1;
while (!queue.isEmpty()) {
Sq curr = (Sq) queue.elementAt(0);
queue.removeElementAt(0);
int i = curr.i;
int j = curr.j;
if (i == r  1 && j == c  1) {
queue.removeAllElements();
return curr.dist;
}
if (i < r  1 && grid[i + 1][j] == 0) {
queue.addElement(new Sq(i + 1, j, curr.dist + 1));
grid[i + 1][j] = 1;
}
if (j < c  1 && grid[i][j + 1] == 0) {
queue.addElement(new Sq(i, j + 1, curr.dist + 1));
grid[i][j + 1] = 1;
}
if (i > 0 && grid[i  1][j] == 0) {
queue.addElement(new Sq(i  1, j, curr.dist + 1));
grid[i  1][j] = 1;
}
if (j > 0 && grid[i][j  1] == 0) {
queue.addElement(new Sq(i, j  1, curr.dist + 1));
grid[i][j  1] = 1;
}
}
return 1;
}
Note that the queue above is actually Vector (has some sync overhead, but is still not that bad). We can't use any other builtin types on OJ (except Hastable), so it is kind of fun.
From what I gathered, it takes 0.2 secs per grid! (Not at my home PC, those 19900 path grids take the longest  0.01sec, otherwise, it can do ~1000 empty 200x200 in sec)
I tried inserting points that are closer to the end in front of the queue(sort of A*), but that didn't change it much.
I used exactly the same BFS algorithm to fill in 10267, I didn't run out of time there.
Can I improve it somehow? Or is it just OJ Java thing?
Darko
10977  Enchanted Forest
I don't know why i'm getting RTE for this prob.(there is no chance of array overrun or negetive indexing,though my memory management is not that efficient) It made me really mad during the contest. Still now i can't find the problem . Plaeaaz somebody help.
Shihab
Code: Select all
Cut
Last edited by shihabrc on Tue Jan 17, 2006 11:09 pm, edited 1 time in total.
Still getting RTE
Sorry. I've changed the size of que to 40100, which is large enough to hold 200x200 points.(according to the prob statement the forest will be at most 200x200 grid.) But i'm still getting RTE. Pleaz help me.
Shihab
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Or only enqueue when it's not in the queue, which is good enough.
Check out http://www.algorithmist.com !
I got WA.ayon wrote:the total area blocked by jigglipuff is circular area. you have to find very carefully the intersections she blocked.
I may be misunderstanding about the dangerous area covered by Jigglypuff.
Then, I make a following test case.
Is my idea wrong?
Code: Select all
10 10
4
1 5
2 3
4 1
8 10
4
3 2 0
5 6 3
10 1 4
10 8 1
0 0
'0' is safe area, '1' is dangerous area or blocked position.
0000100000
0010010000
0100111000
1001111100
0011111110
1001111100
1100111000
1110010001
1111000100
1111101110
So, I think the answer is 24.
Thank you.