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 =  + ;
 int Stack1[MAX_SIZE], Stack2[MAX_SIZE];
 int out[MAX_SIZE];
 int pointer = ;

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

     int j = ;
     for(int i = ; 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 ;
         }

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

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

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

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

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