[UVA][遞迴] 645 - File Mapping
File Mapping
| File Mapping |
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
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
Miguel A. Revilla
2000-01-17
題目描述：
按照 stack 的方式給定資料夾關係，而子資料夾按照輸入順序輸出，
一般文件則以字典順序排序後，放在所有子資料夾後面輸出。
題目解法：
再次見到遞迴處理的奧妙了
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
string s;
void dfs(string name, int dep) {
vector<string> file;
int i;
for(i = 0; i < dep; i++)
printf("| ");
cout << name << endl;
while(s != "]" && s != "*") {
if(s[0] == 'f')
file.push_back(s);
else {
string dir;
dir = s;
cin >> s;
dfs(dir, dep+1);
}
cin >> s;
}
sort(file.begin(), file.end());
for(vector<string>::iterator it = file.begin();
it != file.end(); it++) {
for(i = 0; i < dep; i++)
printf("| ");
cout << *it << endl;
}
}
int main() {
int cases = 0;
while(cin >> s && s != "#") {
if(cases) puts("");
printf("DATA SET %d:\n", ++cases);
dfs("ROOT", 0);
}
return 0;
}