字典树 Ancient Printer

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

 

 

Author

iSea @ WHU

 

 

Source

2010 ACM-ICPC Multi-University Training Contest(3)——Host by WHU

读题很明显是个tire树 敲键盘的次数 = n(每个单词要print一次) + 2 * cnt(每个单词打了要删去) - Maxstr(最后一个最长的不用删去)

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define  LL long long
#define  ULL unsigned long long
#define mod 10007
#define INF 0x7ffffff
#define mem(a,b) memset(a,b,sizeof(a))
#define MODD(a,b) (((a%b)+b)%b)
using namespace std;
const int maxn = 5e5 + 5;
const int NUM = 1e6 + 5;
int n,cnt;
int temp[maxn][30];
void buildTree(char *str)
{
    int tot = 0;
    int len = strlen(str);
    for(int i = 0; i < len; i++){
      int c = str[i] - 'a';
      if(!temp[tot][c]) temp[tot][c] = ++cnt;
      tot = temp[tot][c];
    }
    //colo[tot]++;
}
void init()
{
    mem(temp,0);
    cnt = 0;
}
int query(int Max){return 2 * cnt - Max + n;}
int main()
{
    while(~scanf("%d",&n)){
      char str[55];
      int Max = -1;
      init();
      for(int i = 0; i < n; i++){
        scanf("%s",str);
        int len = strlen(str);
        Max = max(Max,len);
        buildTree(str);
      }
      printf("%d\n",query(Max));

    }


    return 0;
}

 

无语了,数组开大一点就MLE。1e6 和 5e5这种也扣内存。。。。。。行吧,还是题刷少了

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