2016ACM

ACM"/>

2016ACM

【题目传送门】

1383 : The Book List

描述

The history of Peking University Library is as long as the history of Peking University. It was build in 1898. At the end of year 2015, it had about 11,000 thousand volumes of books, among which 8,000 thousand volumes were paper books and the others were digital ones. Chairman Mao Zedong worked in Peking University Library for a few months as an assistant during 1918 to 1919. He earned 8 Dayang per month there, while the salary of top professors in Peking University is about 280 Dayang per month.

Now Han Meimei just takes the position which Chairman Mao used to be in Peking University Library. Her first job is to rearrange a list of books. Every entry in the list is in the format shown below:

CATEGORY 1/CATEGORY 2/..../CATEGORY n/BOOKNAME

It means that the book BOOKNAME belongs to CATEGORY n, and CATEGORY n belongs to CATEGORY n-1, and CATEGORY n-1 belongs to CATEGORY n-2...... Each book belongs to some categories. Let's call CATEGORY1  "first class category", and CATEGORY 2 "second class category", ...ect. This is an example:

MATH/GRAPH THEORY
ART/HISTORY/JAPANESE HISTORY/JAPANESE ACIENT HISTORY
ART/HISTORY/CHINESE HISTORY/THREE KINDOM/RESEARCHES ON LIUBEI
ART/HISTORY/CHINESE HISTORY/CHINESE MORDEN HISTORY
ART/HISTORY/CHINESE HISTORY/THREE KINDOM/RESEARCHES ON CAOCAO

Han Meimei needs to make a new list on which the relationship between books and the categories is shown by indents. The rules are:

1) The n-th class category has an indent of  4×(n-1) spaces before it.
2) The book directly belongs to the n-th class category has an indent of  4×n spaces before it.
3) The categories and books which directly belong to a category X should be list below X in dictionary order. But all categories go before all books. 
4) All first class categories are also list by dictionary order.

For example, the book list above should be changed into the new list shown below:

ARTHISTORYCHINESE HISTORYTHREE KINDOMRESEARCHES ON CAOCAORESEARCHES ON LIUBEICHINESE MORDEN HISTORYJAPANESE HISTORYJAPANESE ACIENT HISTORY
MATHGRAPH THEORY

Please help Han Meimei to write a program to deal with her job.

输入

There are no more than 10 test cases.
Each case is a list of no more than 30 books, ending by a line of "0". 
The description of a book contains only uppercase letters, digits, '/' and spaces, and it's no more than 100 characters.
Please note that, a same book may be listed more than once in the original list, but in the new list, each book only can be listed once. If two books have the same name but belong to different categories, they are different books.

输出

For each test case, print "Case n:" first(n starts from 1), then print the new list as required.

样例输入

B/A
B/A
B/B
0
A1/B1/B32/B7
A1/B/B2/B4/C5
A1/B1/B2/B6/C5
A1/B1/B2/B5
A1/B1/B2/B1
A1/B3/B2
A3/B1
A0/A1
0

样例输出

Case 1:
BAB
Case 2:
A0A1
A1BB2B4C5B1B2B6C5B1B5B32B7B3B2
A3B1

字典树大模拟。因为输出要按字典序输出，所以输入的时候需要给字符串排个序。而且输出的时候在相同层有后继的优先输出（题目没说清QAQ），所以在输出上要一点小处理。

  1 #include<cstdio>
  2 #include<iostream>
  3 #include<cstring>
  4 #include<algorithm>
  5 #define clr(x) memset(x,0,sizeof(x))
  6 #define maxnode 3010
  7 using namespace std;
  8 struct node
  9 {
 10     int lt,rt;
 11     char *str;
 12     int val;
 13 };
 14 struct trie
 15 {
 16     node poi[maxnode];
 17     int nodelen;
 18     int head;
 19     trie() {nodelen=1; head=0; clr(poi);}
 20     void clear() {nodelen=1;head=0; clr(poi);}
 21     void insert(int fa,int now,char *stri)
 22     {
 23 //       printf("fat:%d p:%d nodelen:%d head:%d char:%s\n",fa,now,nodelen,head,stri);
 24         int end=false,i=0;
 25         while(stri[i] && stri[i]!='/')
 26             i++;
 27         if(stri[i]==0)
 28             end=true;
 29         stri[i]=0;
 30         if(!now)
 31         {
 32             now=newnode(stri);
 33             poi[fa].lt=now;
 34             if(!end)
 35                 insert(now,0,stri+i+1);
 36             else
 37                 poi[now].val++;
 38             return ;
 39         }
 40         int q;
 41         while(now && strcmp(poi[now].str,stri)!=0)
 42         {
 43             q=now;
 44             now=poi[now].rt;
 45         }
 46         if(!now)
 47         {
 48            now=newnode(stri);
 49            poi[q].rt=now;
 50         }
 51         if(!end)
 52         {
 53             insert(now,poi[now].lt,stri+i+1);
 54         }
 55         else
 56         {
 57             poi[now].val++;
 58         }
 59         return ;
 60     }
 61     int newnode(char *stri)
 62     {
 63         if(!head)
 64         {
 65             head=nodelen;
 66         }
 67         poi[nodelen].str=stri;
 68         return nodelen++;
 69     }
 70     void output(int node,int dep)
 71     {
 72         if(node==0)
 73         {
 74             return ;
 75         }
 76         int q=node;
 77         while(q)
 78         {
 79             if(poi[q].lt!=0)
 80             {
 81                 for(int i=0;i<dep;i++)
 82                     printf("    ");
 83                 printf("%s",poi[q].str);
 84                 printf("\n");
 85                 output(poi[q].lt,dep+1);
 86             }
 87             q=poi[q].rt;
 88         }
 89         q=node;
 90         while(q)
 91         {
 92             if(poi[q].val)
 93             {
 94                 for(int i=0;i<dep;i++)
 95                     printf("    ");
 96                 printf("%s",poi[q].str);
 97                 printf("\n");
 98             }
 99             q=poi[q].rt;
100         }
101      return ;
102     }
103 
104 
105 }tried;
106 bool cmp(char *a,char *b)
107 {
108     return strcmp(a,b)<0;
109 }
110 char s[110],deal[maxnode][110];
111 char *dir[maxnode];
112 int main()
113 {
114     int n,kase=0;
115     while(fgets(s,110,stdin)!=NULL)
116     {
117         clr(deal);
118         n=0;
119         s[strlen(s)-1]='\0';
120         tried.clear();
121         strcpy(deal[n++],s);
122         dir[0]=deal[0];
123         while(fgets(s,110,stdin)!=NULL && strcmp(s,"0\n")!=0)
124         {
125             s[strlen(s)-1]='\0';
126             strcpy(deal[n],s);
127             dir[n]=deal[n];
128             n++;
129         }
130         sort(dir+0,dir+n,cmp);
131         for(int i=0;i<n;i++)
132         {
133             tried.insert(0,tried.head,dir[i]);
134         }
135         printf("Case %d:\n",++kase);
136         tried.output(tried.head,0);
137     }
138     return 0;
139 }

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