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.
InputThere 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.
OutputFor 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
代码：
1 /*
2 这一道题主要是贪心，最后保留的肯定是长度最长的字符串。
3 如果是最后打印机中也不要剩余一个字母的话，那么就是有创建的节点个数乘与2再加上n(n个打印字符串)个操作
4 最后可以剩余字符，那肯定是打印完最后一个字符串后就结束，那也就是说减少的就是这个最后打印出字符串的长度
5 那肯定这个字符串长度越大那就结果就越小
6
7 之后每创建一个节点还要考虑删除它的操作，再加上要打印n个字符串。最后减去那个最长字符串长度就好了
8
9 */
10 #include <iostream>
11 #include <cstdio>
12 #include <cstring>
13 #include <cstdlib>
14 #include <algorithm>
15 using namespace std;
16 typedef long long ll;
17 const int maxn=26;
18 const int mod=998244353;
19 typedef struct Trie* TrieNode;
20 int result;
21 struct Trie
22 {
23 int sum;
24 TrieNode next[maxn];
25 Trie()
26 {
27 sum=0;
28 memset(next,NULL,sizeof(next));
29 }
30 };
31 void inserts(TrieNode root,char s[55])
32 {
33 TrieNode p = root;
34 int len=strlen(s);
35 for(int i=0; i<len; ++i)
36 {
37 int temp=s[i]-‘a‘;
38 if(p->next[temp]==NULL) p->next[temp]=new struct Trie(),result++;
39 p->next[temp]->sum+=1;
40 p=p->next[temp];
41 }
42 }
43 void Del(TrieNode root)
44 {
45 for(int i=0 ; i<2 ; ++i)
46 {
47 if(root->next[i])Del(root->next[i]);
48 }
49 delete(root);
50 }
51
52 int main()
53 {
54 int n,ans=0,len;
55 char s[55];
56
57 while(~scanf("%d",&n))
58 {
59 ans=0;
60 TrieNode root = new struct Trie();
61 result=0;
62 for(int i=1; i<=n; ++i)
63 {
64 scanf("%s",s);
65 inserts(root,s);
66 len=strlen(s);
67 if(len>ans)
68 {
69 ans=len;
70 }
71 }
72 printf("%d\n",result*2+n-ans);
73 Del(root);
74 }
75
76 return 0;
77 }
原文地址：https://www.cnblogs.com/kongbursi-2292702937/p/12001434.html