poj 1057 FILE MAPPING(递归)_it is often helpful for computer users to see a vi-优快云博客

本文介绍了一个程序，该程序能够读取一系列表示计算机文件结构的数据集，并将这些数据集以可视化的形式展示出来。每个数据集包括文件和目录名称，程序会递归地遍历这些目录并按字母顺序列出所有子目录和文件。

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FILE MAPPING

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 2564		Accepted: 1213

Description

It is often helpful for computer users to see a visual representation of the file structure on their computers. The "explorer" in Microsoft Windows is an example of such a system. Before the days of graphical user interfaces, however, such visual representations were not possible. The best that could be done was to show a static "map"of directories and files, using indentation as a guide to directory contents. For example:

ROOT

|	DIR1

|        File1

|	File2

|	File3

|	DIR2

|	DIR3

|	File1

File1

File2

This shows that the root directory contains two files and three subdirectories. The first subdirectory contains 3 files, the second is empty and the third contains one file.

Input

Write a program that reads a series of data sets representing a computer file structure. A data set ends with a line containing a single *, and the end of valid data is denoted by a line containing a single #. The data set contains a series of file and directory names. (The root directory is assumed to be the starting point.) The end of a directory is denoted by a ']' Directory names begin with a lower case 'd' and file names begin with a lower case 'f' File names may or may not have an extension (such as fmyfile.dat or fmyfile). File and directory names may not contain spaces.

Output

Note that the contents of any directory should list any subdirectories first, followed by files, if any. All files should be in alphabetical order within each directory. Note that each data set output is marked by the label "DATA SET x:" where x denotes the number of the set, beginning at 1. Note also the blank line between the output data sets. Each level of indentation should show a '|' followed by 5 spaces.

Sample Input

file1
file2
dir3
dir2
file1
file2
]
]
file4
dir1
]
file3
*
file2
file1
*
#

Sample Output

DATA SET 1:
ROOT
|     dir3
|     |     dir2
|     |     file1
|     |     file2
|     dir1
file1
file2
file3
file4

DATA SET 2:
ROOT
file1
file2

tips:一点都不喜欢做这样的题目，输入搞得我很乱。

ss每次输入的字符按串，因为每次到dir时才会有递归，dir为dir文件名

s临时字符串变量，用于临时代替ss的值，已更新ss字符串

#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<vector>
using namespace std;
string ss;
void dfs(int depth,string dir)
{
	for(int i=0;i<depth;i++)cout<<"|     ";
	vector<string>file;
	cout<<dir<<endl;string s;
	while(true)
	{
		if(ss[0]==']'||ss[0]=='*')break;
		else if(ss[0]=='f')file.push_back(ss);
		else s=ss,cin>>ss,dfs(depth+1,s);
		cin>>ss;
	} 
	sort(file.begin(),file.end());
	for(int i=0;i<file.size();i++)
	{
		for(int i=0;i<depth;i++)cout<<"|     ";
		cout<<file[i]<<endl;
	}
}
int main()
{
	for(int i=1;cin>>ss,ss!="#";i++)
	{
		cout<<"DATA SET "<<i<<":"<<endl;
		dfs(0,"ROOT");
		cout<<endl;
	}
	
	return 0;
 }