JUMPPY - Jumppy and the Grid

Jumppy likes to jump! One day Jumppy went to a park. Jumppy can jump all over the park. The park can be thought of as a square grid with square cells of side length 1. The contents of the grid is either 0 (zero) or X. There are certain things which Jumppy likes. They are:

  • Jumppy likes rectangles.
  • Jumppy likes X.

Therefore Jumppy will jump only in the rectangles consisting of X. A rectangle is defined as follows:

  1. The whole rectangular region should contain only X.
  2. The rectangle should be surrounded with 0 or the boundary of the grid.
  3. The diagonally adjacent cell (see the definition) of the corner of the rectangle may be X or 0. (Refer to the first example).

Diagonally adjacent cell: Suppose the given cell has coordinates (p, q) then its diagonally adjacent cells would have coordinates (p+1, q+1), (p+1, q-1), (p-1, q+1), (p-1, q-1).

Now Jumppy is curious how many rectangles are there in the park. Help Jumppy find the number of rectangle.

Input

An integer n denoting the size of the grid. Then n lines follow each containing a string of n characters describing the square grid. All the characters will be either 0 or X.

Output

Output the number of rectangles in the given grid.

Constraints

0 < n <= 1000

Examples:

Input:
4
XX00
XX00
00XX
00XX

Output:
2
Input:
5
00000
0XXX0
0XXX0
0XXX0
00X00

Output:
0
Input:
5
00000
0XXX0
0X0X0
0XXX0
00000

Output:
0
Input:
3
X0X
0X0
X0X

Output:
5

Explanation

Case 1: As can be seen there are two rectangles as highlighted.

Case 2: The grid contains no rectangles because it violates the second condition of the definition.

Case 3: The grid contains no rectangles because it violates the first condition of the definition.

Case 4: The individual X in this case can be considered as rectangles.


Added by:Anant Kumar
Date:2014-11-03
Time limit:1s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: ASM64

hide comments
2016-02-06 21:18:51
Didn't experience any problems here, and O(n^2) without optimizing constants passed in 0.19
2015-07-29 20:41:19 Rishav Goyal
@author : after trying hard on this problem, i was getting run time error. then i tried to upload one AC code of my friend , its giving Runtime error. Either something is wrong on spoj or input files. author u can verify what i said.
2015-01-29 20:35:40 :D
I don't know if time limit was changes, but currently O(N^2) with getchar map reading and O(N) memory complexity easily passes.
2014-11-08 17:34:44 newbie
time limit strict for O(n^2) to pass
2014-11-08 17:34:44 mehmetin
O(n^2) works fine without fast io.
2014-11-08 17:34:44 Akhilesh Anandh
Time limit too strict. O(n^2) algorithm is giving TLE even with fast IO. (Note that reading in the input itself is O(n^2), so you can't do better than that.)
There was a problem in input file. Fixed it and your solution got accepted.

Edit: Thanks

Last edit: 2014-11-07 18:46:15
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