HDOJ-1702 栈和队列问题[或模拟]

ACboy needs your help again!

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1697    Accepted Submission(s): 906

Problem Description

ACboy was kidnapped!!
he miss his mother very much and is very scare now.You can't image how dark the room he was put into is, so poor :(.
As a smart ACMer, you want to get ACboy out of the monster's labyrinth.But when you arrive at the gate of the maze, the monste say :" I have heard that you are very clever, but if can't solve my problems, you will die with ACboy."
The problems of the monster is shown on the wall:
Each problem's first line is a integer N(the number of commands), and a word "FIFO" or "FILO".(you are very happy because you know "FIFO" stands for "First In First Out", and "FILO" means "First In Last Out").
and the following N lines, each line is "IN M" or "OUT", (M represent a integer).
and the answer of a problem is a passowrd of a door, so if you want to rescue ACboy, answer the problem carefully!

 

 

Input

The input contains multiple test cases.
The first line has one integer,represent the number oftest cases.
And the input of each subproblem are described above.

 

 

Output

For each command "OUT", you should output a integer depend on the word is "FIFO" or "FILO", or a word "None" if you don't have any integer.

 

 

Sample Input

4 4 FIFO IN 1 IN 2 OUT OUT 4 FILO IN 1 IN 2 OUT OUT 5 FIFO IN 1 IN 2 OUT OUT OUT 5 FILO IN 1 IN 2 OUT IN 3 OUT

 

 

Sample Output

1 2 2 1 1 2 None 2 3

 

 

Source

2007省赛集训队练习赛（1）

 

 

Recommend

lcy

 

 

 

code1：栈和队列的使用

 1 #include<iostream>

 2 #include<stack>

 3 #include<queue>

 4 using namespace std;

 5 int main()

 6 {

 7     char str[12];

 8     int n,m,cases;

 9     scanf("%d",&n);

10     while(n--)

11     {

12         scanf("%d%str",&cases,str);

13         if(str[2]=='F')                   //队列

14         {

15             queue<int>Que;                       //创建一个int型队列

16             while(cases--)

17             {

18                 scanf("%s",str);

19                 if(str[0]=='I')

20                 {

21                     scanf("%d",&m);

22                     Que.push(m);                   //入队

23                 }

24                 else if(!Que.empty())

25                 {

26                     printf("%d\n",Que.front());

27                     Que.pop();                     //出队

28                 }

29                 else

30                     printf("None\n");

31             }

32         }

33         else

34         {

35             stack<int>Sta;                         //创建一个int型的栈

36             while(cases--)

37             {

38                 scanf("%s",str);

39                 if(str[0]=='I')

40                 {

41                     scanf("%d",&m);

42                     Sta.push(m);

43                 }

44                 else if(!Sta.empty())

45                 {

46                     printf("%d\n",Sta.top());

47                     Sta.pop();

48                 }

49                 else

50                     printf("None\n");

51             }

52         }

53     }

54     return 0;

55 }

---

 

code2：模拟

 1 #include<iostream>

 2 using namespace std;

 3 

 4 int data[10002];

 5 char str[5];

 6 char ord[5];

 7 

 8 int main()

 9 {

10     int n;

11     int i;

12     int cases;

13     int cnt;

14     int num;

15     int front;

16     while(~scanf("%d",&n))

17     {

18         while(n--)

19         {

20             memset(data,0,sizeof(data));

21             scanf("%d%s",&cases,str);

22             if(str[2]=='L')

23             {

24                 cnt=0;

25                 for(i=0;i<cases;i++)

26                 {

27                     scanf("%s",ord);

28                     if(ord[0]=='I')

29                     {

30                         scanf("%d",&num);

31                         data[cnt]=num;

32                         cnt++;

33                     }

34                     if(ord[0]=='O')

35                     {    

36                         if(cnt>0)

37                         {

38                             printf("%d\n",data[cnt-1]);

39                             cnt--;

40                         }

41                         else

42                             printf("None\n");

43                     }

44                 }

45             }

46             else

47             {

48                 cnt=0;

49                 front=0;

50                 for(i=0;i<cases;i++)

51                 {

52                     scanf("%s",ord);

53                     if(ord[0]=='I')

54                     {

55                         scanf("%d",&num);

56                         data[cnt]=num;

57                         cnt++;

58                     }

59                     if(ord[0]=='O')

60                     {

61                         if(front<cnt)

62                         {

63                             printf("%d\n",data[front]);

64                             front++;

65                         }

66                         else

67                             printf("None\n");

68                     }

69                 }

70             }

71         }

72     }

73     return 0;

74 }

75 /*

76 4

77 7 FIFO

78 IN 1

79 IN 3

80 IN 5

81 OUT

82 OUT

83 OUT

84 OUT

85 */