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:

ART    HISTORY        CHINESE HISTORY            THREE KINDOM                RESEARCHES ON CAOCAO                RESEARCHES ON LIUBEI            CHINESE MORDEN HISTORY        JAPANESE HISTORY            JAPANESE ACIENT HISTORYMATH    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/AB/AB/B0A1/B1/B32/B7A1/B/B2/B4/C5A1/B1/B2/B6/C5A1/B1/B2/B5A1/B1/B2/B1A1/B3/B2A3/B1A0/A10

样例输出

Case 1:B    A    BCase 2:A0    A1A1    B        B2            B4                C5    B1        B2            B6                C5            B1            B5        B32            B7    B3        B2A3
    B1
#include <iostream>#include <cmath>#include <vector>#include <cstdlib>#include <cstdio>#include <climits>#include <ctime>#include <cstring>#include <queue>#include <stack>#include <list>#include <algorithm>#include <map>#include <set>#define LL long long#define Pr pair<int,int>#define fread(ch) freopen(ch,"r",stdin)#define fwrite(ch) freopen(ch,"w",stdout)using namespace std;const int INF = 0x3f3f3f3f;const int mod = 1e9+7;const double eps = 1e-8;const int maxn = 11234;struct Node{    map<string,int>nex;    set<string> hav;} arr[maxn];int _cnt;int newNode(){    arr[_cnt].nex.clear();    arr[_cnt].hav.clear();    return _cnt++;}string spi[maxn];void insert(int st,int dep,int len){    if(dep == len - 1){        arr[st].hav.insert(spi[dep]);        return;    }    if(arr[st].nex.find(spi[dep]) == arr[st].nex.end())        arr[st].nex[spi[dep]] = newNode();    insert(arr[st].nex[spi[dep]],dep+1,len);}void insert(int root,string v){    int len = 0;    spi[len] = "";    for(auto x : v)    {        if(x != '/')        {            spi[len] = spi[len] + string(1,x);        }        else        {            len++;            spi[len] = "";        }    }    len++;    insert(root,0,len);}void out(int st,int deep,string v){    if(deep != -1)    {        for(int i=0; i<4*(deep); i++)            cout<<" ";        cout<<v<<endl;    }    for(auto x : arr[st].nex)    {        out(x.second,deep+1,x.first);    }    for(auto x : arr[st].hav){         for(int i=0; i<4*(deep + 1); i++)            cout<<" ";        cout<<x<<endl;    }}int main(){    //fread("in.in");    //fwrite("in.in");    string inp = "";    int icase = 1;    do    {        _cnt = 0;        int root = newNode();        if(inp[0] != '0')        {            if(inp != "")                insert(root,inp);            while(getline(cin,inp) && inp != "0")            {                insert(root,inp);            }        }        cout<<"Case "<<icase++<<":"<<endl;        out(root,-1,"");    }    while(getline(cin,inp));    return 0;}