Broken Keyboard（模拟数组或者双重链表的运用）

这题我是大写的服气，辛辛苦苦搞了个双重链表结果还一直不对，不对就算了，书上源代码打进去还是不对，我能怎么办我也很无奈。不过这题还是让我对双重链表更加了解和运用了！还是可以的！

You’re typing a long text with a broken keyboard. Well it’s not so badly broken. The only problem with the keyboard is that sometimes the “home” key or the “end” key gets automatically pressed (internally). You’re not aware of this issue, since you’re focusing on the text and did not even turn on the monitor! After you finished typing, you can see a text on the screen (if you turn on the monitor). In Chinese, we can call it Beiju.

Your task is to find the Beiju text. Input There are several test cases. Each test case is a single line containing at least one and at most 100,000 letters, underscores and two special characters ‘[’ and ‘]’. ‘[’ means the “Home” key is pressed internally, and ‘]’ means the “End” key is pressed internally.

The input is terminated by end-of-file (EOF).

Output For each case, print the Beiju text on the screen.

Sample Input This_is_a_[Beiju]_text

[[]][][]Happy_Birthday_to_Tsinghua_University

Sample Output

BeijuThis_is_a__text

Happy_Birthday_to_Tsinghua_University

附上我的代码加正确答案

  1 #include<iostream>
  2 #include<set>
  3 #include<cstring>
  4 #include<cstdio>
  5 #include<cmath>
  6 #include<algorithm>
  7 using namespace std;
  8 struct date
  9 {
 10     char ch;
 11     date *next;
 12 };
 13 class myqueue
 14 {
 15 public:
 16 
 17     date *top,*tail,*mid;
 18     myqueue()
 19     {
 20         top=NULL;
 21         tail=NULL;
 22     }
 23     void inserts(char x)
 24     {
 25         date *p;
 26         p=new date;
 27         p->ch=x;
 28         p->next=NULL;
 29         if(top==NULL)
 30         {
 31             top=p;
 32             tail=p;
 33         }
 34         else
 35         {
 36             tail->next=p;
 37             tail=p;
 38 
 39 
 40         }
 41     }
 42     void insertm(char x)
 43 {
 44 
 45     date *p;
 46     p=new date;
 47     p->ch=x;
 48     p->next=mid->next;
 49     mid->next=p;
 50     mid=p;
 51 }
 52     void insertq(char x)
 53     {
 54 
 55         date *p;
 56         p=new date;
 57         p->ch=x;
 58         p->next=NULL;
 59         if(top==NULL)
 60         {
 61              top=p;
 62             tail=p;
 63 
 64 
 65         }
 66         else
 67         {
 68             p->next=top;
 69             top=p;
 70 
 71         }
 72         mid=top;
 73    }
 74    void clears()
 75    {
 76        top=tail=NULL;
 77     }
 78     void print()
 79     {
 80         date *p=top;
 81         while(p!=NULL)
 82         {
 83             cout<<p->ch;
 84             p=p->next;
 85 
 86         }
 87 }
 88 
 89 };
 90 int main()
 91 {
 92     char x[100005];
 93     myqueue s;
 94     while(cin>>x)
 95     {
 96         s.clears();
 97         int flag=1;
 98         int len=strlen(x);
 99         for(int i=0;i<len;i++)
100         {
101             if(x[i]=='[')
102                 flag=0;
103             else if(x[i]==']')
104                     flag=1;
105                     else
106                     {
107                         if(flag==1)
108                             s.inserts(x[i]);
109                         else if(flag==0)
110                             {
111                                 s.insertq(x[i]);
112                                 flag=2;
113                            }
114                          else  if(flag==2)
115                             s.insertm(x[i]);
116 
117   }
118 
119 }
120           s.print();
121           cout<<endl;
122 
123  }
124 
125     return 0;
126 
127 
128 }

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 using namespace std;
 5 #define N 100005
 6 int main()
 7 {
 8     char s[N];
 9     int last,cur,next[N];
10     while(scanf("%s",s+1)==1)
11     {
12         int n=strlen(s+1);
13         last=cur=0;
14         next[0]=0;
15         for(int i=1;i<=n;i++)
16         {
17             char ch=s[i];
18             if(ch=='[')
19                 cur=0;
20             else if(ch==']')
21                 cur=last;
22             else
23             {
24                 next[i]=next[cur];
25                 next[cur]=i;
26                 if(cur==last)
27                     last=i;
28                 cur=i;
29             }
30         }
31         for(int i=next[0];i!=0;i=next[i])
32             printf("%c",s[i]);
33         printf("
");
34     }
35     return 0;
36 }

Broken Keyboard（模拟数组或者双重链表的运用）

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