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 21 2 3 5 4

Output

pushpoppushpoppushpoppushpushpoppop

Example 2

Input

5 53 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！

1 #include 
       
         2 #include 
        
          3 #include 
         
           4 using namespace std; 5  6 const int MAX_SIZE = 1600000 + 10; 7 int Stack1[MAX_SIZE], Stack2[MAX_SIZE]; 8 int out[MAX_SIZE]; 9 int pointer = 0;10 11 void push(int a)12 {13     pointer++;14     Stack1[pointer] = a;15     Stack2[pointer] = a;16 }17 18 void pop()19 {20     pointer--;21 }22 23 int top1()24 {25     return Stack1[pointer];26 }27 28 int top2()29 {30     return Stack2[pointer];31 }32 33 int main()34 {35     int n, m;36     scanf("%d %d", &n, &m);37 38     for(int i = 1; i <= n; ++i)39     {40         scanf("%d", &out[i]);41     }42 43     int j = 0;44     for(int i = 1; i <= n; ++i)45     {46         ///3个if 1个while 就模拟了栈混洗过程47         if(out[i] < top1())48         {49             printf("No");50             return 0;51         }52 53         while(j < out[i])54         {55             push(++j);56             printf("push(%d)\n", j);57         }58 59         if(m < pointer)60         {61             printf("No");62             return 0;63         }64 65         if(out[i] == top1())66         {67             printf("pop\n");68             pop();69         }70     }71 72     j = 0;73     for(int i = 1; i <= n; ++i)74     {75         while(j < out[i])76         {77             push(++j);78             puts("push");79         }80 81         if(out[i] == top2())82         {83             pop();84             puts("pop");85         }86     }87     return 0;88 }