hiho 1383 The Book List（windows文件系统）

 The Book List

Time Limit: 1000ms

Case Time Limit: 1000ms

Memory Limit: 256MB

Description

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.

Input

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.

Output

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

Sample Input

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

Sample Output

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

题意：给你多个文件的目录（有可能有相同的文件），让你排序后输出

注意：文件夹中的文件要按照windows的排列方式，即所有的文件夹先显示在前面，再显示文件，文件夹与文件各自都要按照字典序排序

相同的文件只要输出一次

思路：明显的字典树，可是不会处理各种字符串，后面看了别人的思路才懂。 

注意map和set一样是会自动把里面的元素按照字典序排序的。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
using namespace std;
struct Node
{
    map<string,int> ma;
    map<string,int> ma1;
} p[3050];
void dfs(int u,int deep)
{
    for(map<string, int>::iterator iter = p[u].ma.begin(); iter != p[u].ma.end();++iter)
    {
        for(int i=1;i<=4*deep;i++)
            printf(" ");
        cout<<iter->first<<endl;
        dfs(iter->second,deep+1);
    }
    for(map<string,int> ::iterator iter=p[u].ma1.begin();iter!=p[u].ma1.end();++iter)
    {
        for(int i=1;i<=4*deep;i++)
            printf(" ");
        cout<<iter->first<<endl;
    }
}
int main()
{
    char a[3050];
    int num=0,Tcase=1;
    for(int i=0;i<3050;i++) p[i].ma.clear(),p[i].ma1.clear();
    while(gets(a))
    {
        int len=strlen(a);
        if(len==1&&a[0]=='0')
        {
            printf("Case %d:\n",Tcase++);
            dfs(0,0);
            for(int i=0;i<3050;i++) p[i].ma.clear(),p[i].ma1.clear();
            num=0;
            continue;
        }
        int cnt=0;
        a[len]='@';
        string m="";
        for(int i=0; i<=len; i++)
        {
            if(a[i]=='/')
            {
                if(p[cnt].ma.find(m)==p[cnt].ma.end())
                {
                    p[cnt].ma[m]=++num;
                    cnt=num;
                    m="";
                }
                else
                {
                    cnt=p[cnt].ma[m];
                    m="";
                }
            }
            else if(a[i]=='@')
            {
                if(p[cnt].ma1.find(m)!=p[cnt].ma1.end()) continue;
                p[cnt].ma1[m]=++num;
            }
            else m+=a[i];
        }
    }
    return 0;
}