CF 25 E 三个字符串 KMP模板

Time Limit: 2000MS

Memory Limit: 262144KB

64bit IO Format: %I64d & %I64u

Description

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 s_i. 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.

Sample Input

Input

ab
bc
cd

Output

Input

abacaba
abaaba
x

Output

Source

Codeforces Beta Round #25 (Div. 2 Only)

题意：将3个字符串连接起来，重复的可以重叠不用计算，输出最短的长度

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 100000+20;
char s[4][maxn];
int next[4][maxn],len[5];

void get_next(int i)
{
    int k=-1,j=0;
    next[i][0]=-1;
    while(j<len[i])
       if(k==-1||s[i][j]==s[i][k])
         {
            k++;
            j++;
            next[i][j]=k;
         }
       else k=next[i][k];
}

int kmp(int a,int b)
{
     int  i=0,j=0;
     while(j<len[b]&&i<len[a])
     if(i==-1||s[a][i]==s[b][j])
        {
            i++;j++;
        }
     else i=next[a][i];
     return i;
}

int main()
{
    while(~scanf("%s %s %s",s[0],s[1],s[2]))
    {
        int ans=0;
        len[0]=strlen(s[0]);
        len[1]=strlen(s[1]);
        len[2]=strlen(s[2]);
        for(int i=0;i<3;i++) get_next(i);

        for(int i=0;i<3;i++)
          for(int j=0;j<3;j++)
              {
                  if(i==j) continue;
                  for(int k=0;k<3;k++)
                  {
                      if(k==i||k==j) continue;
                      int x=kmp(i,j);
                      int y1=kmp(k,i);
                      int y2=kmp(k,j);
                      ans=max(ans,max(x+y1,x+y2));
                  }
              }
        printf("%d
",len[0]+len[1]+len[2]-ans);
    }
    return 0;
}

　　分析：只要求出三个字符串所能得到的最大匹配数就好，