北京赛区(2016)网络赛 题目1 : The Book List 【字典树】

题目1 : The Book List

时间限制:1000ms
单点时限:1000ms
内存限制:256MB

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:

ART
HISTORY
CHINESE HISTORY
THREE KINDOM
RESEARCHES ON CAOCAO
RESEARCHES ON LIUBEI
CHINESE MORDEN HISTORY
JAPANESE HISTORY
JAPANESE ACIENT HISTORY
MATH
GRAPH 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:
B
A
B
Case 2:
A0
A1
A1
B
B2
B4
C5
B1
B2
B6
C5
B1
B5
B32
B7
B3
B2
A3
B1

题解：

先插入字典树，然后对字典树上的元素进行排序；

排序遵守两个原则：

1. 类别名优先于书名。

2. 字典序小的优先于字典序大的。

注意：

结束是“一个” 0

单个换行可能有影响---

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
struct trie{
int lei;
char ch[101];
trie * next[33];
}boot,PP[3000],op[50];
int kp,lp;
int ssscmp(trie * A,trie * B)
{
if (A->lei!=B->lei)
return A->lei-B->lei;
return strcmp(A->ch,B->ch);
}
void ADD(int ll)
{
trie * kk=&boot;
for (int i=0;i<ll;i++)
{
bool fafe=true;
int j;
for (j=0;;j++)
{
if (kk->next[j]==NULL)
break;
if (ssscmp(kk->next[j],&op[i])==0)
{
fafe=false;
kk=kk->next[j];
break;
}
}
if (fafe)
{
strcpy(PP[kp].ch,op[i].ch);
PP[kp].lei=op[i].lei;
kk->next[j]=&PP[kp];
kk=kk->next[j];
kp++;
}
}
}
void pan(trie * xx)
{
trie * c;
for (int i=0;xx->next[i]!=NULL;i++)
{
for (int j=i+1;xx->next[j]!=NULL;j++)
{
if (ssscmp(xx->next[i],xx->next[j])>0)
{
c=xx->next[i];
xx->next[i]=xx->next[j];
xx->next[j]=c;
}
}
}
for (int i=0;xx->next[i]!=NULL;i++)
pan(xx->next[i]);
}
void output(trie * xx,int ko)
{
if (ko!=-1)
{
for (int i=0;i<ko;i++)
{
for (int j=0;j<4;j++)
printf(" ");
}
printf("%s\n",xx->ch);
}
for (int i=0;xx->next[i]!=NULL;i++)
{
output(xx->next[i],ko+1);
}
}
int main()
{
char ch[101];
kp=0;
int ca=1;
while (gets(ch))
{
if (ch[0]==0)
continue;
if (ch[0]=='0'&&strlen(ch)==1)
{
pan(&boot);
printf("Case %d:\n",ca++);
output(&boot,-1);
kp=0;
memset(PP,0,sizeof(PP));
memset(op,0,sizeof(op));
for (int i=0;i<33;i++)
boot.next[i]=NULL;
}
else
{
lp=0;int kl=0;
int ll=strlen(ch);
memset(op,0,sizeof(op));
for (int i=0;i<=ll;i++)
{
if (ch[i]=='/'||ch[i]==0)
{
if (ch[i]=='/')
op[lp].lei=0;
else
op[lp].lei=1;
lp++;
kl=0;
}
else
op[lp].ch[kl++]=ch[i];
}
ADD(lp);
}
}
}