Problem:


Score : 1500 points

Problem Statement

Problem F and F2 are the same problem, but with different constraints and time limits.

We have a board divided into N horizontal rows and N vertical columns of square cells.
The cell at the i-th row from the top and the j-th column from the left is called Cell (i,j).
Each cell is either empty or occupied by an obstacle.
Also, each empty cell has a digit written on it.
If A_{i,j}= 1, 2, ..., or 9, Cell (i,j) is empty and the digit A_{i,j} is written on it.
If A_{i,j}= #, Cell (i,j) is occupied by an obstacle.

Cell Y is reachable from cell X when the following conditions are all met:

Cells X and Y are different.

Cells X and Y are both empty.

One can reach from Cell X to Cell Y by repeatedly moving right or down to an adjacent empty cell.

Consider all pairs of cells (X,Y) such that cell Y is reachable from cell X.
Find the sum of the products of the digits written on cell X and cell Y for all of those pairs.

Constraints

1 \leq N \leq 1500

A_{i,j} is one of the following characters: 1, 2, ... 9 and #.

Input

Input is given from Standard Input in the following format:

N
A_{1,1}A_{1,2}...A_{1,N}
A_{2,1}A_{2,2}...A_{2,N}
:
A_{N,1}A_{N,2}...A_{N,N}

Output

Print the sum of the products of the digits written on cell X and cell Y for all pairs (X,Y) such that cell Y is reachable from cell X.

Sample Input 1

2
11
11

Sample Output 1

5

There are five pairs of cells (X,Y) such that cell Y is reachable from cell X, as follows:

X=(1,1), Y=(1,2)

X=(1,1), Y=(2,1)

X=(1,1), Y=(2,2)

X=(1,2), Y=(2,2)

X=(2,1), Y=(2,2)

The product of the digits written on cell X and cell Y is 1 for all of those pairs, so the answer is 5.

Sample Input 2

4
1111
11#1
1#11
1111

Sample Output 2

47

Sample Input 3

10
76##63##3#
8445669721
75#9542133
3#285##445
749632##89
2458##9515
5952578#77
1#3#44196#
4355#99#1#
#298#63587

Sample Output 3

36065

Sample Input 4

10
4177143673
7#########
5#1716155#
6#4#####5#
2#3#597#6#
6#9#8#3#5#
5#2#899#9#
1#6#####6#
6#5359657#
5#########

Sample Output 4

6525



Sample input:
2
11
11


Sample output:
5

Submission status: Time Limit Exceeded

Solution code:

def elongation(edges1, edges2, data, total):
    new_edges = list()
    for e1 in edges1:
        for e2 in edges2:
            if e1[1] == e2[0]:
                if [e1[0], e2[1]] not in new_edges:
                    new_edges.append([e1[0], e2[1]])
                    total += \
                        data[e1[0] // n][e1[0] % n] * data[e2[1] // n][e2[1] % n]
            else:
                continue
    return new_edges, total


n = int(input().strip())

data = list()

total = 0

for i in range(n):
    data.append([
        int(x)
        if x != '#'
        else x
        for x in list(input().strip())])

edges = [[]]

for i in range(n - 1):
    for j in range(n):
        if data[i][j] != '#' and data[i + 1][j] != '#':
            edges[0].append([
                i * n + j,
                (i + 1) * n + j])
            total += data[i][j]*data[i + 1][j]
        else:
            continue

for i in range(n):
    for j in range(n - 1):
        if data[i][j] != '#' and data[i][j + 1] != '#':
            edges[0].append([
                i * n + j,
                i * n + j + 1])
            total += data[i][j]*data[i][j + 1]
        else:
            continue

for i in range(2 * n - 3):
    new_edges, total = elongation(edges[i], edges[0], data, total)
    edges.append(new_edges)

print(total)

Submission status: Runtime Error

Solution code:
import copy
n = int(input())
a = [[i for i in input().split()] for i in range(n)] 

total = 0
turnable = True
for i in range(n):
    for j in range(n):
        if a[i][j] == "#":
          continue

        x = 1
        y = 0
        copie_map = copy.deepcopy(a)
        sum_list = []
        first_i = i
        first_j = j

        while True:
            i += x
            j += y
            if copie_map[i][j] == "#":
                if turnable == False:
                    if i == first_i and j == first_j:
                       total += sum(sum_list)*a[i][j]
                       break
                    sum_list.append(int(b[i][j]))
                    copie_map[i-x][j-y] = "#"
                i -= x
                j -= y
                x, y = y, x
                turnable = False
            else:
                turnable = True
print(total)