Train dispatcher (Train)

Train dispatcher (Train)

---

Description

Figure 1 shows the structure of a station for train dispatching.

Figure 1

In this station, A is the entrance for each train and B is the exit. S is the transfer end. All single tracks are one-way, which means that the train can enter the station from A to S, and pull out from S to B. Note that the overtaking is not allowed. Because the compartments can reside in S, the order that they pull out at B may differ from that they enter at A. However, because of the limited capacity of S, no more that m compartments can reside at S simultaneously.

Assume that a train consist of n compartments labeled {1, 2, …, n}. A dispatcher wants to know whether these compartments can pull out at B in the order of {a1, a2, …, an} (a sequence). If can, in what order he should operate it?

Input

Two lines:

1st line: two integers n and m;

2nd line: n integers separated by spaces, which is a permutation of {1, 2, …, n}. This is a compartment sequence that is to be judged regarding the feasibility.

Output

If the sequence is feasible, output the sequence. “Push” means one compartment goes from A to S, while “pop” means one compartment goes from S to B. Each operation takes up one line.

If the sequence is infeasible, output a “no”.

Example 1

Input

5 2
1 2 3 5 4

Output

push
pop
push
pop
push
pop
push
push
pop
pop

Example 2

Input

5 5
3 1 2 4 5

Output

No

Restrictions

1 <= n <= 1,600,000

0 <= m <= 1,600,000

Time: 2 sec

Memory: 256 MB

1. Principles and elements: simple application stack. Initially empty stack, loop through the stack order of the array.
   - if the current top of the stack should be smaller than the elements of the stack, the stack number is sequentially behind the recording stack
   - if the current top element of the stack should equal elements, out of the stack, the pointer shift through the array, recording the stack
   - if the current is greater than the top element of the stack should the elements, then the order of the stack is not possible, the output No, and then the program ends
2. Challenges: None
3. Time and space complexity: time complexity \ (O (n) \) , the space complexity \ (O (n) \)

#include "iostream"
#include "cstdio"

using namespace std;
const int maxn = 4e6 + 10;
int a[maxn];
int st[maxn];

const int SZ = 1<<20;  //快速io
struct fastio{
    char inbuf[SZ];
    char outbuf[SZ];
    fastio(){
        setvbuf(stdin,inbuf,_IOFBF,SZ);
        setvbuf(stdout,outbuf,_IOFBF,SZ);
    }
}io;
int main() {
    int n, m;
    scanf("%d %d", &n, &m);
    int now = 1;
    int top = 0, tot = 0;
    int x;
    st[0] = -1;
    for (int i = 0; i < n; i++) {
        scanf("%d", &x);
        while (now <= x) {
            st[++top] = now++;
            a[tot++] = 0;
        }
        if (top > m) {
            printf("No\n");
            return 0;
        }
        if (st[top] == x) {
            top--;
            a[tot++] = 1;
        } else {
            printf("No\n");
            return 0;
        }
    }
    for (int i = 0; i < tot; i++) {
        if (a[i]) {
            printf("pop\n");
        } else {
            printf("push\n");
        }
    }
    return 0;
}