Algorithm competition entry classic 6.1.2 stack and queue-rails, algorithm competition entry classic rails

There is a railway station in a city, and the tracks are laid as shown below. There are n cars entering the station from the direction A, and they are numbered 1 ~ n in the order of arrival. Your task is to get them into the tracks in the B direction and get out of the station in a certain order. In order to reorganize the cars, you can use the transit station C; this is a station that can park any number of cars, but because the end is capped, the cars entering C must exit C in the reverse order. For each compartment, once you enter C from A, you cannot return to A; once you enter B from C, you cannot return to C. In other words, at any time, there are only two options: A-> C and C-> B.
Sample input:
5
1 2 3 4 5
5
5 4 1 2 3
6
6 5 4 3 2 1
Sample output:
Yes
No
Yes


 1 #include <stdio.h>
 2 #define MAXN 1000 + 10
 3 int n, target [MAXN];
 4
 5 int main (void)
 6 {
 7 while (scanf ("% d", & n) == 1)
 8     {
 9 int stack [MAXN], top = 0;
10 int A = 1, B = 1;
11 for (int i = 1; i <= n; i ++)
12 scanf ("% d", & target [i]);
13 int ok = 1;
14 while (B <= n)
15 {
16 if (A == target [B]) {A ++; B ++;} // The car enters and leaves the transit station C in order, then it exits the loop
17 else if (top && stack [top] == target [B]) {top--; B ++;} /// If the car enters the transit station C in reverse order, it will jump out of the loop
18 else if (A <= n) stack [++ top] = A ++; // Adjust the carriages to reverse transfer station C
19 else {ok = 0; break;} // the carriages are neither entering nor leaving the transit station C in order or in reverse order
20}
21 printf ("% s \ n", ok? "Yes": "No");
twenty two     }
23 return 0;
twenty four }
View Code
analysis:
栈 1. Stack: In the transit station C, the carriage conforms to the principle of last-in, first-out, called the stack, which is the LIFO table; where LIFO stands for Last In First Out. Since only one end of the stack grows, you only need an array stack and a pointer to the top of the stack (which always points to the top element of the stack).
2. For the second input case:
B <= n A <= n stack [++ top] = A ++;
1 <= 5 1 <= 5 stack [1] = 1; top = 1; A = 2;
           2 <= 5 stack [2] = 2; top = 2; A = 3;
            3 <= 5 stack [3] = 3; top = 3; A = 4;
            4 <= 5 stack [4] = 4; top = 4; A = 5;
            5 <= 5 stack [5] = 5; top = 5; A = 6;
Top && stack [top] == target [B] top--; B ++;
5 && 5 == 5 top = 4; B = 2;
4 && 4 == 4 top = 3; B = 3;
3 && 3 == 1 Out of the while loop
3. For convenience, the array index starts from 1. For example: target [1] refers to the number of the first car in the target sequence, and stack [1] refers to the bottom element of the stack (the stack is empty if and only if top = 0).

C ++ provides a simpler way of processing-STL queue:


 1 #include <cstdio>
 2 #include <stack>
 3 using namespace std;
 4 #define MAXN 1000 + 10
 5 int n, target [MAXN];
 6
 7 int main (void)
 8 {
 9 while (scanf ("% d", & n) == 1)
10 {
11 stack <int> s;
12 int A = 1, B = 1;
13 for (int i = 1; i <= n; i ++)
14 scanf ("% d", & target [i]);
15 int ok = 1;
16 while (B <= n)
17 {
18 if (A == target [B]) {A ++; B ++;} // The car enters and leaves the transit station C in order, then the loop is out
19 else if (! S.empty () && s.top () == target [B]) {s.pop (); B ++;} // If the car enters the transit station C in reverse order, it will jump out of the loop
20 else if (A <= n) s.push (A ++); // Adjust the carriage to reverse transfer station C
21 else {ok = 0; break;} // The carriages are neither entering nor leaving the transit station C in order or in reverse order
twenty two         }
23 printf ("% s \ n", ok? "Yes": "No");
twenty four     }
25 return 0;
26}
View Code
Reprinted at: https://www.cnblogs.com/zhuangwei/p/5352781.html