列车调度(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

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 <cstdio>
 2 #include <iostream>
 3 #include <cstring>
 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 }