Codeforces Beta Round #25 (Div. 2 Only)

E. Test

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Sometimes it is hard to prepare tests for programming problems. Now Bob is preparing tests to new problem about strings — input data to his problem is one string. Bob has 3 wrong solutions to this problem. The first gives the wrong answer if the input data contains the substring s1, the second enters an infinite loop if the input data contains the substring s2, and the third requires too much memory if the input data contains the substring s3. Bob wants these solutions to fail single test. What is the minimal length of test, which couldn't be passed by all three Bob's solutions?

Input

There are exactly 3 lines in the input data. The i-th line contains string si. All the strings are non-empty, consists of lowercase Latin letters, the length of each string doesn't exceed 105.

Output

Output one number — what is minimal length of the string, containing s1, s2 and s3 as substrings.

Examples

input

abbccd

output

4

input

abacabaabaabax

output

11

题意：

输入三个字符串，问你存在着三个字符串子串的字符串的最短长度。

POINT：

枚举A(3,3)种可能。

特殊情况若字符串包含，则处理一下。

#include <iostream>#include <stdio.h>#include <algorithm>#include <math.h>using namespace std;const int maxn = 1e5+55;#define LL long longint nxt[maxn];void prenxt(string s){int i=0,j;nxt[0]=j=-1;while(i<s.length()){while(j!=-1&&s[j]!=s[i]) j=nxt[j];nxt[++i]=++j;}}int kmp(string s1,string s2){prenxt(s2);int i=0,j=0;while(i<s1.length()){while(j!=-1&&s1[i]!=s2[j]) j=nxt[j];i++,j++;if(j==s2.length()) return (int)s2.length();}return j;}int ans=0x3f3f3f3f;void doit(string s1, string s2,string s3){int len=kmp(s1,s2);string s="";s+=s1;for(int i=len;i<s2.length();i++){s+=s2[i];}len=kmp(s,s3);ans=min(ans,(int)(s.length()-len+s3.length()));}int main(){string s1,s2,s3;cin>>s1>>s2>>s3;doit(s1,s2,s3);doit(s1,s3,s2);doit(s2,s1,s3);doit(s3,s1,s2);doit(s2,s3,s1);doit(s3,s2,s1);printf("%d\n",ans);}