Code for implementing the C ++ train entry Algorithm

Code for implementing the C ++ train entry Algorithm

This article mainly introduces the implementation code of the C ++ train entry algorithm. For more information, see

[Problem description]

 

There is a railway station in a city where rails are laid. There are n trains entering the station from the direction, numbered 1 ~ N. Your task is to let them enter the Track B in a specific order and drive out of the station. To restructure the carriage, you can use Transfer Station C. This is a station that can park any number of trains, but because of the end cap, the carriage that enters C must be driven out in reverse order. For each carriage, once it is moved from A to C, it cannot return to A. Once it is moved from C to B, it cannot return to C. In other words, there are only two options at any time: A → C and C → B.

 

This problem is similar to the train entry in the previous data structure experiment and is simpler. I tried to write it myself, but there is still a big gap between the amount of code for the reference answers in the book. The Code is as follows:

 

 

The Code is as follows:

# Include <iostream>

Using namespace std;

Const int MAXSIZE = 100;

Void main ()

{

Int n;

Cin> n;

Int a [MAXSIZE], B [MAXSIZE];

Int stack [MAXSIZE];

For (int I = 0; I <n; I ++)

{

A [I] = I + 1;

Cin> B [I]; // The Order of the output stack.

}

Int top =-1;

Int count = 0;

Int I = 0;

For (;;)

{

If (I <n)

{

++ Top;

Stack [top] = a [I ++]; // inbound stack

Cout <"PUSH" <endl;

}

 

If (stack [top] = B [count])

{

Top --; count ++;

Cout <"POP" <endl;

}

Else if (I = n)

{

Cout <"NO" <endl;

Break;

}

If (count = n)

{

Cout <"YES" <endl;

Break;

}

If (top <-1)

{

Cout <"NO" <endl;

Break;

}

}

 

}

 

 

 

The reference code is as follows:

 

 

 

The Code is as follows:

# Include <iostream>

Using namespace std;

Const int MAXN = 1000 + 10;

Int n, target [MAXN];

Void main ()

{

While (cin> n)

{

Int stack [MAXN], top = 0;

Int A = 1, B = 1; // A is used to record the number of incoming stacks, and B is used to record the train number of the derailed train

For (int I = 1; I <= n; I ++)

Cin> target [I]; // record the derailment Sequence

Int OK = 1;

While (B <= n)

{

If (A = target [B]) {A ++; B ++ ;}

Else if (top & stack [top] = target [B]) {top --; B ++;} // output stack

Else if (A <= n) stack [++ top] = A ++; // inbound stack

Else {OK = 0; break ;}

}

If (OK)

Cout <"Yes" <endl;

Else

Cout <"No" <endl;

}

 

 

Similarly, STL can be used for implementation. You only need to make minor changes to the answers referenced in the book. The Code is as follows:

 

 

 

The Code is as follows:

/* Implement the STL stack */

# Include <iostream>

# Include <stack>

Using namespace std;

Const int MAXN = 1000 + 10;

Int n, target [MAXN];

Int main ()

{

While (cin> n)

{

Stack <int> s;

Int A = 1, B = 1;

For (int I = 1; I <= n; I ++)

Cin> target [I];

Int OK = 1;

While (B <= n)

{

If (A = target [B]) {A ++; B ++ ;}

Else if (! S. empty () & s. top () = target [B]) {s. pop (); B ++ ;}

Else if (A <= n) s. push (A ++ );

Else {OK = 0; break ;}

}

If (OK)

Cout <"YES" <endl;

Else

Cout <"NO" <endl;

}

}

 

 

[Summary]

 

There is still room for optimization for self-written code. Learn the clear logic expression of the reference answer. Learn how to use the STL stack.

 

In the data structure experiment, you should add a judgment on the restrictions on the train's inbound rail.

 

Beginners of C ++, who do not know how deep they are, are currently taking a step by step in the "water question" phase, and crossing the river by feeling the stones and aligning them with Daniel.