Write a C++ program called coin_collection.cpp to collect maximum number of coins on an n x m board with inaccessible cells. For details, read the question 5 of the textbook page 291.

Your program should ask a user for a text filename to read a 2-D array of a sequence of numbers. In the array, 1 indicates that theres a coin on the cell, while 0 means no coin. A cell with the number 2 indicates that the cell is not accessible. Your program should display maximum possible coins and path to collect it. If you have more than one path, you only need to display one. For the homework, you can assume that the board size is less than or equal to 25 x 25. When you write the program, dont forget to include Title, Abstract, ID (A four-digit number), Name, and Date.

This is a sample run of the program on the cloud9.

$ g++ -o coin_collection coin_collection.cpp

$ ./coin_collection

Enter a file name: ./t1.txt

Max coins: 3

Path:(1,1)->(2,1)->(2,2)->(3,2)->(3,3)->(4,3)->(4,4)

For the sample run, t1.txt has the following context:

4 4

0 2 1 1

0 1 2 1

0 0 1 2

1 2 0 1

The first line (= 4 and 4 in the example) indicates the numbers of rows and columns in the file. One blank space is used to delimiter the data. Note that theres no blank space at the end of each line. If your program does not read the file properly, your program will get no credit. In the sample execution, the starting indexes of the path are 1, not 0.

This is another sample run.

$ g++ -o coin_collection coin_collection.cpp

$ ./coin_collection

Enter a file name: ./t2.txt

Max coins: 1

Path:(1,1)->(1,2)->(1,3)->(1,4)->(2,4)

For the sample run, t2.txt has the following context:

2 4

0 1 0 0

1 1 2 0

How would you modify the dynamic programming algorithm for the coincollecting problem if some cells on the board are inaccessible for the robot? Apply your algorithm to the board below, where the inaccessible cells are shown by 2s. How many optimal paths are there for this board?

int F[n][m];

F[1][1] = board[1][1];

for (int j = 2; j <= m; j++) {

F[1][j] = F[1][j-1] + board[1][j];

}

for (int i = 2; i <= n; i++) {

F[i][1] = F[i-1][1] + board[i][1];

for (int j = 2; j <= m; j++) {

F[i][j] = max(F[i-1][j], F[i][j-1]) + board[i][j];

}

}

return F[n][m];