Ancient Printer（tire树）

Ancient Printer

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1511    Accepted Submission(s):
748

Problem Description

The contest is beginning! While preparing the contest,
iSea wanted to print the teams' names separately on a single
paper.
Unfortunately, what iSea could find was only an ancient printer: so
ancient that you can't believe it, it only had three kinds of
operations:

● 'a'-'z': twenty-six letters you can type
● 'Del': delete
the last letter if it exists
● 'Print': print the word you have typed in the
printer

The printer was empty in the beginning, iSea must use the three
operations to print all the teams' name, not necessarily in the order in the
input. Each time, he can type letters at the end of printer, or delete the last
letter, or print the current word. After printing, the letters are stilling in
the printer, you may delete some letters to print the next one, but you needn't
delete the last word's letters.
iSea wanted to minimize the total number of
operations, help him, please.

 

Input

There are several test cases in the input.

Each
test case begin with one integer N (1 ≤ N ≤ 10000), indicating the number of
team names.
Then N strings follow, each string only contains lowercases, not
empty, and its length is no more than 50.

The input terminates by end of
file marker.

 

Output

For each test case, output one integer, indicating
minimum number of operations.

 

Sample Input

2

freeradiant

freeopen

 

Sample Output

21

Hint

The sample's operation is:
f-r-e-e-o-p-e-n-Print-Del-Del-Del-Del-r-a-d-i-a-n-t-Print

题目大意：类似于把这些单词在电脑上都打一遍，输入一个单词后需要删除，最后一个不需要删除，问需要至少敲多少个按键

想到tire树就好写了。

用tire树构造所有单词，构造时申请了多少个node内存就输出了多少个字母，因为还要删除，所以需要乘以2，但我们会把最长的单词放在最后输入，然后省去了删除操作，所以需要加上他的strlen()，然后加上单词个数=num_Print;

 /*******************************

 Date    : 2015-12-09 22:15:29
 Author  : WQJ (1225234825@qq.com)
 Link    : http://www.cnblogs.com/a1225234/
 Name     : 

 ********************************/
 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cmath>
 #include <cstring>
 #include <string>
 #include <set>
 #include <vector>
 #include <queue>
 #include <stack>
 #define LL long long
 using namespace std;
 struct node
 {
     node* next[];
     node()
     {
         for(int i=;i<;i++)
             next[i]=NULL;
     }
 };
 int n;
 char a[];
 int ans;
 node* root;
 void insert(char* s,int len)
 {
     int i,j;
     node* p=root;
     for(i=;i<len;i++)
     {
         int pos=s[i]-'a';
         if(p->next[pos]==NULL)
         {
             ans++;
             p->next[pos]=new node;
             p=p->next[pos];
         }
         else
             p=p->next[pos];
     }
     return;
 }
 int main()
 {
     freopen("in.txt","r",stdin);
     while(scanf("%d",&n)!=EOF)
     {
         int i,j;
         int Max=;
         ans=;
         root=new node;
         for(i=;i<n;i++)
         {
             scanf("%s",a);
             int len=strlen(a);
             Max=max(Max,len);
             insert(a,len);
         }
         printf("%d\n",ans*+n-Max);
     }
     return ;
 }