标签:put directly close single complete while view cin pos
C - Petya and File System
Time Limit:3000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Submit Status Practice CodeForces 66C
Appoint description:
Description
Recently, on a programming lesson little Petya showed how quickly he can create files and folders on the computer. But he got soon fed up with this activity, and he decided to do a much more useful thing. He decided to calculate what folder contains most subfolders (including nested folders, nested folders of nested folders, and so on) and what folder contains most files (including the files in the subfolders).
More formally, the subfolders of the folder are all its directly nested folders and the subfolders of these nested folders. The given folder is not considered the subfolder of itself. A file is regarded as lying in a folder, if and only if it either lies directly in this folder, or lies in some subfolder of the folder.
For a better understanding of how to count subfolders and files for calculating the answer, see notes and answers to the samples.
You are given a few files that Petya has managed to create. The path to each file looks as follows:
diskName:\folder1\folder2\...\ foldern\fileName
It is also known that there is no file whose path looks like diskName:\fileName. That is, each file is stored in some folder, but there are no files directly in the root. Also let us assume that the disk root is not a folder.
Help Petya to find the largest number of subfolders, which can be in some folder, and the largest number of files that can be in some folder, counting all its subfolders.
Input
Each line of input data contains the description of one file path. The length of each line does not exceed 100, and overall there are no more than 100 lines. It is guaranteed, that all the paths are correct and meet the above rules. It is also guaranteed, that there are no two completely equal lines. That is, each file is described exactly once.
There is at least one line in the input data.
Output
Print two space-separated numbers. The first one is the maximal number of possible subfolders in a folder (including nested folders, nested folders of nested folders, and so on). The second one is the maximal number of files in a folder (including nested files in subfolders). Note that the disks are not regarded as folders.
Sample Input
Input
C:\folder1\file1.txt
Output
0 1
Input
C:\folder1\folder2\folder3\file1.txt
C:\folder1\folder2\folder4\file1.txt
D:\folder1\file1.txt
Output
3 2
Input
C:\file\file\file\file\file.txt
C:\file\file\file\file2\file.txt
Output
4 2
Hint
In the first sample we have one folder on the "C" disk. It has no subfolders, which is why the first number in the answer is 0. But this folder contains one file, so the second number of the answer is 1.
In the second sample we have several different folders. Consider the "folder1" folder on the "C" disk. This folder directly contains one folder, "folder2". The "folder2" folder contains two more folders — "folder3" and "folder4". Thus, the "folder1" folder on the "C" drive has exactly 3 subfolders. Also this folder contains two files, even though they do not lie directly in the folder, but they are located in subfolders of "folder1".
In the third example we see that the names of some folders and some subfolders are identical. Consider the "file" folder, which lies directly on the "C" disk. That folder contains another "file" folder, which in turn contains another "file" folder, which contains two more folders, "file" and "file2". Thus, the "file" folder, which lies directly on the "C" disk, contains 4 subfolders.
题意:
给定一些文件的路径,要求统计出包含最多子目录的目录的包含子目录数量以及包含文件最多的目录的包含文件数量。
分析:
利用map存储路径信息(包含多少子目录和多少文件),每一行都进行倒序遍历即可。
1 #include <cstdio> 2 #include <string> 3 #include <map> 4 #include <iostream> 5 using namespace std; 6 string str; 7 map<string,int> FILE_RECORDER[10],FOLDER_RECORDER[10]; 8 int main(){ 9 int max1 = 0,max2 = 0; 10 while(cin >> str){ 11 int DISK = str[0] - ‘C‘; 12 int FOLDERNUM = 0; 13 for(int i = str.length() ; i >= 3 /*不考虑根目录*/; i--)if(str[i] == ‘\\‘){ 14 string now = str.substr(0,i); 15 FOLDER_RECORDER[DISK][now] += FOLDERNUM; 16 // each file is described exactly once. 17 if(FILE_RECORDER[DISK][now] == 0) FOLDERNUM++; // 是新文件 18 FILE_RECORDER[DISK][now]++; 19 max1 = max(max1,FOLDER_RECORDER[DISK][now]); 20 max2 = max(max2,FILE_RECORDER[DISK][now]); 21 } 22 } 23 cout << max1 << " " << max2 << endl; 24 return 0; 25 }
标签:put directly close single complete while view cin pos
原文地址:http://www.cnblogs.com/cyb123456/p/6045390.html