POJ 3856 deltree（模拟）

Description

You have just run out of disk space and decided to delete some of your directories. Rationally, you will first have an exploration of what you have in your file system. And more rationally, you will do this exploration through a command line interface. The interface used in this problem is called “MSDOS--”, since it is something like MSDOS with fewer features. The commands of MSDOS-- are as follows: 
1. cd <directory>  Assuming <directory> to be the name of a relative descendant of current directory, this command changes the current directory to <directory>. For example, when the current directory is “\A\B\” and one of its descendants is “C\D”, the execution of “cd C\D” will change the current directory to “\A\B\C\D\”. 
2. cd \  This command changes the current directory to “\” (the root of the file system). For example, when the current directory is “\A\B\”, the execution of “cd \” will change the current directory to “\”. 
3. cd ..  Assuming the current directory to be anything except “\”, this command changes the current directory to its parent directory. For example, when the current directory is “\A\B\”, the execution of “cd ..” will change the current directory to “\A\”. 
4. cd \<directory>  This command is equivalent to the execution of the following two commands:  cd \  cd <directory> 
5. dir  This command lists the name of files and directories directly in the current directory, each on a separate line.  These file/directory names are made up of (lowercase and uppercase) letters, digits, and dots (“.”). Directory names precede the file names in the list, and each one, comes alone in a single line. On the contrary, each file name is accompanied by its size separated by a space. A sample output of “dir” is as follows:  HW1  HW1.old  Syllab.pdf 10000  notes.txt 3241 
6. deltree <directory>  Assuming <directory> to be the name of a relative descendant of current directory, this command tries to delete <directory> and all its descendant files and subdirectories (and thus, freeing that much of space). For example, when the current directory is “\A\B\” and one of its descendants is “C\D”, the execution of “deltree C\D” will try to delete directory “\A\B\C\D\” and all of its descendant files and directories. 
7. deltree \<directory>  This command is equivalent to the execution of the following two commands:  cd \  deltree <directory> 
8. exit  This command terminates the command line interface. 
A “scenario” is an exploration (a consistent series of “cd” and “dir” commands and their results, starting from root) followed by exactly one “deltree” command. Given a scenario, you are to find the maximum space guaranteed to be freed by executing its “deltree” command.

Input

Input contains multiple independent scenarios. There is an empty line after each scenario. The input ends with an “exit” command. There is a “>” sign before each command in the input (with no spaces in between). The length of each file name does not exceed 50. You may assume that the input is correct.

Output

Write the result of the ith scenario as a single integer on the ith line of output.

题目大意：模拟一个CMD的运行，假定所有给定的语句都是正确的。

思路：丧心病狂模拟题系列。注意细节，比如我用一个目录dir两次，不要同一个文件算两次，再如有A\B，我删掉了B，然后再删A的时候，不要再把B的容量给算上了。我觉得这题样例还算有良心，我过了样例就AC了o(╯□╰)o

PS：我的代码虽然暴力是暴力了点，不过丧心病狂模拟题的重点，不是要快，而是要好写，好调，准确……

代码（0MS）：

 #include <cstdio>
 #include <iostream>
 #include <cstring>
 #include <algorithm>
 using namespace std;

 const int MAXN = ;

 struct Node {
     char name[];
     int next, pre, siz;
     bool isFile, del;
 };

 Node a[MAXN];
 int head[MAXN], ecnt, root, now;
 char s[], tmp[];

 void clear() {
     root = now = ;
     head[root] = ;
     ecnt = ;
 }

 int new_sub(int cur, char *name, int size, bool isFile) {
     a[ecnt].pre = cur;
     strcpy(a[ecnt].name, name);
     a[ecnt].siz = size;
     a[ecnt].isFile = isFile;
     a[ecnt].next = head[cur];
     a[ecnt].del = ;
     head[ecnt] = ;
     return head[cur] = ecnt++;
 }

 int to_sub(int cur, char *name, int siz = , int isFile = ) {
     for(int p = head[cur]; p; p = a[p].next)
         if(strcmp(a[p].name, name) == ) return p;
     return new_sub(cur, name, siz, isFile);
 }

 int get_pre(int cur) {
     return a[cur].pre;
 }

 int get_root() {
     return root;
 }

 void _strcpy(char *&src, char *tar) {
     int len = ;
     while(*src != '\\' && *src != ) tar[len++] = *src, ++src;
     tar[len] = ;
     if(*src == '\\') ++src;
 }

 void cd(char *s) {
     if(*s == '\\') now = get_root(), ++s;
     while(*s != ) {
         _strcpy(s, tmp);
         now = to_sub(now, tmp);
     }
 }

 int str_to_num(char *s) {
     int ret = ;
     for(int i = ; s[i]; ++i)
         ret = ret *  + s[i] - '';
     return ret;
 }

 void dir() {
     while(gets(s) && *s != '>') {
         int i = ;
         for(i = ; s[i]; ++i)
             if(s[i] == ' ') break;
         if(s[i] != ' ') {
             to_sub(now, s);
         }
         else {
             s[i] = ;
             to_sub(now, s, str_to_num(s + i + ), );
         }
     }
 }

 int dfs_del(int cur) {
     if(a[cur].del) return ;
     a[cur].del = true;
     int ret = ;
     for(int p = head[cur]; p; p = a[p].next) {
         if(a[p].isFile) ret += a[p].siz;
         else ret += dfs_del(p);
     }
     return ret;
 }

 void deltree(char *s) {
     if(*s == '\\') now = root, ++s;
     int cur = now;
     while(*s != ) {
         _strcpy(s, tmp);
         cur = to_sub(cur, tmp);
     }
     printf("%d\n", dfs_del(cur));
 }

 int main() {
     clear();
     gets(s);
     while(strcmp(s, ">exit") != ) {
         if(s[] == ) {//next exploration
             clear();
             gets(s);
         }
         if(strcmp(s, ">cd ..") == ) {
             now = get_pre(now);
             gets(s);
             continue;
         }
         if(strncmp(s, ">cd", ) == ) {
             char *name = s + ;
             while(*name == ' ') ++name;
             cd(name);
             gets(s);
             continue;
         }
         if(strcmp(s, ">dir") == ) {
             dir();
             continue;
         }
         if(strncmp(s, ">deltree", ) == ) {
             char *name = s + ;
             while(*name == ' ') ++name;
             deltree(name);
             gets(s);
             continue;
         }
     }
 }