int r, c;
cin >> r >> c;
int d;
cin >> d;
d++;
int x[r][c];
for (int i = 0; i<=r-1; i++)
for (int j = 0; j<=c-1; j++)
cin >> x[i][j];
long long int ans[r][c];
int temp1, temp2, temp3, temp4;
ans[0][0]=1;
for (int p = 0; p<=r-1; p++)
for (int q = 0; q<=c-1; q++)
{
temp1 = (p>=d)?ans[p-d][q]%20011:0;
temp2 = (q>=d)?ans[p][q-d]%20011:0;
temp3 = (p>=1)?ans[p-1][q]%20011:0;
temp4 = (q>=1)?ans[p][q-1]%20011:0;
if (temp3+temp4-temp1-temp2>=0)
ans[p][q]=temp3+temp4-temp1-temp2;
else
ans[p][q]=0;
if (x[p][q]==0)
ans[p][q]=0;
ans[0][0]=1;
}
cout << ans[r-1][c-1];
return 0;
}
I know of the O(n^3) & the O(n^4) solutions. Given the constraints, they should work… But my solutions is, in fact, O(n^2). The method is called GRAPH CUTTING. It is a very correct algorithm. There are probably slight mistakes in memory allocation, implementation of the algorithm etc. So, I need help in debugging the solution…So please try to…THANKS!!!
This is again a dynamic programming problem. Try thinking of the number of ways to reach a particular (x, y) with a certain number of consecutive moves (for both directions allowed), and how to formulate that in terms of smaller (x, y).
The value stored in ans[p][q] may exceed the integer limits, and that is a possible reason for the ‘wrong answer’. So, you must make sure that it is small, and still get the correct answer, and the best way to do that is :
And this will work as, (a%c + b%c)%c = (a+b)%c, and ans[p][q] will always be inside the limits (0 - 20010).
Also, ans being an int** should be sufficient now.
@arpanb8 My answer, unfortunately, still holds. Try to think a bit more! Think in four dimensions – a particular (x, y), coming from left/right, with 0 to K consecutive moves. If you want the complete answer, say so.
