题目

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

纯模拟即可,可以用树存储,也可以用链表存储,任意一种方法都行

使用 链表 纯模拟即可,需要注意一些细节的处理
优先输出书库,再输出书本,同时应该按照字典序输出

因此,可以用两个不同的链表分别维护子书库和书本

直接使用 STL 即可,单纯考验 STL 的熟练度

代码

/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:oyohyee@oyohyee.com

かしこいかわいい?
エリーチカ!
要写出来Хорошо的代码哦~
*/
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <set>
#include <list>
#include <queue>
#include <stack>
#include <string>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

typedef long long LL;

const int INF = 0x7FFFFFFF;
const double eps = 1e-10;

int kase = 1;

struct CATEGORY {
    list<CATEGORY> *child;
    list<CATEGORY> *book;
    string s;
    CATEGORY(string b) {
        list<CATEGORY> *p = new list<CATEGORY>;
        p->clear();
        child = p;

        p = new list<CATEGORY>;
        p->clear();
        book = p;

        s = b;
    }
    bool operator < (string rhs) {
        return s < rhs;
    }
};

list<CATEGORY> L;

list<CATEGORY>::iterator pushsk(list<CATEGORY>::iterator t,string s) {
    list<CATEGORY>::iterator it = t->child->begin();
    it = lower_bound(t->child->begin(),t->child->end(),s);
    if(it == t->child->end() || it->s != s) {
        t->child->insert(it,CATEGORY(s));
        it--;
    }
    return it;
}

void pushs(list<CATEGORY>::iterator t,string s) {
    list<CATEGORY>::iterator it = t->book->begin();
    it = lower_bound(t->book->begin(),t->book->end(),s);
    if(it == t->book->end() || it->s != s) {
        t->book->insert(it,CATEGORY(s));
        it--;
    }
}

void DFS(list<CATEGORY> *L,int n) {
    if(L == NULL)
        return;
    list<CATEGORY>::iterator it = L->begin();
    while(it != L->end()) {
        for(int i = 0;i < n;i++)
            cout << "    ";
        if(n != -1)
            cout << it->s << endl;
        DFS(it->child,n + 1);
        DFS(it->book,n + 1);

        delete it->child;
        delete it->book;//处理内存泄露问题
        ++it;
    }
}

bool Do() {
    L.clear();
    L.push_back(CATEGORY("MAIN"));//主书库

    string s;
    while(1) {
        if(!(getline(cin,s)))
            return false;
        if(s == "0")
            break;
        size_t len = s.size();
        string t = "";
        list<CATEGORY>::iterator it = L.begin();

        for(size_t i = 0;i < len;i++) {
            if(s[i] == '/') {
                it = pushsk(it,t);
                t = "";
            } else {
                t += s[i];
            }
        }
        pushs(it,t);
    }
    cout << "Case " << kase++ << ":" << endl;
    DFS(&L,-1);
    return true;
}

int main() {
    cin.tie(0);
    cin.sync_with_stdio(false);

    while(Do());

    return 0;
}