hdu 3460 字典树

网址：http://acm.hdu.edu.cn/showproblem.php?pid=3460

Ancient Printer

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 289    Accepted Submission(s): 132

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
   
    
  

 

 

题目大意是要求利用三个基本操作，完成n个队名的输出

借助字典树，可以很好的解决

首先，可以推出，最后输出的肯定是最长的那个字符串...

其余的在字典树中，可以这样表示，每一个结点(根节点除外)，都会出现 进和del 2 次操作

另外还有n个字符串的n次输出

结果就是 2*节点数-最大字串的长度+n

#include <iostream> using namespace std; #include <cstdio> #include <cstring> #include <cstdlib> typedef struct Node { Node *next[26]; Node() { for(int i=0;i<=25;i++) next[i]=NULL; } }node; int ans;// 记录答案 char t[51]; int _biggest;//用来统计最大的那个字符串的长度 void insert(node *&r,char p[]) { node *u=r; if(r==NULL) r=u=new node; int len=strlen(p); if(_biggest<len) _biggest=len; int flag=0; while(flag<len) { int dd=p[flag]-'a'; if(u->next[dd]==NULL) { u->next[dd]=new node; ans++; } u=u->next[dd]; flag++; } return ; } int main() { //freopen("1.in","r",stdin); int n; while(scanf("%d",&n)!=EOF) { _biggest=0; ans=0; node *root=NULL; for(int i=1;i<=n;i++) { scanf("%s",t); insert(root,t); } // 节点 * 2 - 最大的字符串长度 + n次输出操作 ans=ans*2+n-_biggest; printf("%d/n",ans); } return 1; }