列车调度(Train)

列车调度(Train)

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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. 原理与要点：栈的简单应用。初始栈为空，循环遍历出栈顺序的数组。
   - 如果当前栈顶元素小于应该出栈的元素，则顺次把后面的数字入栈，记录入栈
   - 如果当前栈顶元素等于应该出栈的元素，则出栈，遍历数组的指针后移 ，记录出栈
   - 如果当前栈顶元素大于应该出栈的元素，则说明该出栈顺序不可能实现，输出No，然后结束程序
2. 遇到的问题：无
3. 时间和空间复杂度： 时间复杂度(O(n))，空间复杂度(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
");
            return 0;
        }
        if (st[top] == x) {
            top--;
            a[tot++] = 1;
        } else {
            printf("No
");
            return 0;
        }
    }
    for (int i = 0; i < tot; i++) {
        if (a[i]) {
            printf("pop
");
        } else {
            printf("push
");
        }
    }
    return 0;
}