C + + based on stack to realize rail problem _c language

This article describes the C + + based on stack to achieve rail problems. Share to everyone for your reference. The specific analysis is as follows:

The sample diagram looks like this:

Classic Stack problem! The first time to do the thinking is too confusing, now read the Liu Rugia book, rearrange.

#include <stdio.h> #include <string.h>/************************************************************** * Use array A to store the pre-adjusted compartment number sequence, array B to store the adjusted compartment number sequence * Use stack stack to store the compartment in transit c * Use the ix_a to point to the front compartment of a, with ix_b pointing to the front of the carriage * * If Ix_b has pointed to the, then all The carriage has been in the specified order, otherwise * if the Ix_b pointed to the carriage equal to the ix_a-pointed carriage, then directly to the ix_a open to B, otherwise * if the ix_b point of the car is equal to stack elements in stack, then the stack out of the top of the car, otherwise * if there is a car, then the A into the transit station C
, otherwise * output has no solution.
/char stack[1010];
Char a[1010];
Char b[1010];
 int main (int argc, char *argv[]) {scanf ("%s%s", a,b);
 int len_a = strlen (a);
 int len_b = strlen (b);
 /* Initialize stack */int top =-1;
 int tot = 0;
 int ix_a = 0;
 int ix_b = 0;
   while (1) {if (a[ix_a] = = B[ix_b]) {printf ("in\n");
   printf ("out\n");
   ++ix_b;
   ++ix_a;
  if (Ix_b = = len_b)/* The car has been all out, operation success/break;
   else if (tot!= 0/* First check the stack for null/&& B[ix_b] = = Stack[top]) {printf ("out\n");
   --tot;
   --top;
   ++ix_b;
  if (Ix_b = = Len_b) break;
 else if (ix_a!= len_a) {printf ("in\n");
   ++tot;
   ++top;
   Stack[top] = a[ix_a];
   ++ix_a;
  if (Ix_b = = Len_b) break; else {printf (no solution!)
   \ n ");
  return 0; } printf (Finish!
 \ n ");
return 0; }

I hope this article will help you with your C + + programming.