hdu 3460 Ancient Printer

Ancient Printer

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1411    Accepted Submission(s): 705

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

 

Author

iSea @ WHU

 

Source

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

 

有点像贪心～不过感觉很巧妙

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <string>
#include <cmath>
#include <map>
#include <algorithm>
#define N 500015
#define INF 1000000
#define ll long long
using namespace std;
string word[100005];
int check(string a,string b)
{
    int i = 0;
    while(i < (int)a.length() && i < (int)b.length() && a[i] == b[i]) {i++;}
    return a.length() - i + b.length() - i + 1;//删除非共同部分，加上新部分再加上按一下输出
}
int main(void)
{
    int n,i,sum,maxs;
    while(scanf("%d",&n) != -1)
    {
        for(i = 0; i < n; i++)
            cin>>word[i];
        sort(word,word+n);
        sum = 0;
        maxs = word[0].length();
        sum +=  word[0].length() + 1;
        for(i = 1; i < n; i++)
        {
            sum += check(word[i-1],word[i]);
            if((int)word[i].length() > maxs)
                maxs = word[i].length();
        }
        sum += word[n-1].length();
        //先算出所有的单词都敲完并删完所需要的敲击次数
        
        
        sum -= maxs;//最后一个可以不删除，就留下最长的那个
        printf("%d
",sum);
    }
    return 0;
}