Infix to Postfix Write a C++ program that will accept infix expressions (like 5*(4+8)) and convert them to postfix. You are to use Dijkstra's algorithm for converting. Dijkstra's Algorithm: Read in 1 line into a string Print the string while there are characters left to process in the string ----- | get a token (skip over blanks) | if the token is a digit then output(token) | else | ----- | | if the token is '(' then push(token) | | else | | ----- | | | if the token is ')' then | | | ----- | | | | while the top item is not '(' | | | | pop(temp) and output(temp); | | | | pop(temp) | | | ----- | | | else | | | ----- | | | | if the stack is empty then push(token) | | | | else | | | | ----- | | | | | while the stack is not empty | | | | | and the priority (token) <= priority (top item on the stack) | | | | | pop(temp) and output(temp) | | | | | push(token) | | | | ----- | | | ----- | | ---- | ----- ----- while the stack is not empty do pop(temp) and output(temp) Precedence of the Operators: operators : ^ * / + - ( precedence: 3 2 2 1 1 0 where ^ means exponentiation input for the assignment: 2 + 3 * 5 2 + 3 * 5 ^ 6 2 + 3 - 5 + 6 - 4 + 2 - 1 2 + 3 * (5 - 6) - 4 2 * 3 ^ 5 * 6 - 4 (2 + 3) * 6 ^ 2 Output for the assignment 1: 2 + 3 * 5 235*+ 2: 2 + 3 * 5 ^ 6 2356^*+ 3: 2 + 3 - 5 + 6 - 4 + 2 - 1 23+5-6+4-2+1- 4: 2 + 3 * (5 - 6) - 4 2356-*+4- 5: 2 * 3 ^ 5 * 6 - 4 235^*6*4- 6: (2 + 3) * 6 ^ 2 23+62^* You might also try: 7: ( ( ( ( 2 + 3 - 4 ) / 2 + 8 ) * 3 * ( 4 + 5 ) / 2 / 3 + 9 ) ) 23+4-2/8+3*45+*2/3/9+ Programming Notes: You are to write a well-composed program. The stack routines are to be defined as methods in a class and are to be in a separate file. //A sample c++ program to read a file 1 line at a time #include  #include  #include  using namespace std; int main() { string aline; ifstream inData; inData.open("infix.data"); while ( getline(inData,aline) ) { cout << aline << endl; } inData.close(); }