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！

 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 }

View Code