清华学堂 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

感觉很简单的题目，就是难以下手，还是自己不行，加油吧！

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;

const int MAX_SIZE = 1600000 + 10;
int Stack1[MAX_SIZE], Stack2[MAX_SIZE];
int out[MAX_SIZE];
int pointer = 0;

void push(int a)
{
    pointer++;
    Stack1[pointer] = a;
    Stack2[pointer] = a;
}

void pop()
{
    pointer--;
}

int top1()
{
    return Stack1[pointer];
}

int top2()
{
    return Stack2[pointer];
}

int main()
{
    int n, m;
    scanf("%d %d", &n, &m);

    for(int i = 1; i <= n; ++i)
    {
        scanf("%d", &out[i]);
    }

    int j = 0;
    for(int i = 1; i <= n; ++i)
    {
        ///3个if 1个while 就模拟了栈混洗过程
        if(out[i] < top1())
        {
            printf("No");
            return 0;
        }

        while(j < out[i])
        {
            push(++j);
            printf("push(%d)\n", j);
        }

        if(m < pointer)
        {
            printf("No");
            return 0;
        }

        if(out[i] == top1())
        {
            printf("pop\n");
            pop();
        }
    }

    j = 0;
    for(int i = 1; i <= n; ++i)
    {
        while(j < out[i])
        {
            push(++j);
            puts("push");
        }

        if(out[i] == top2())
        {
            pop();
            puts("pop");
        }
    }
    return 0;
}