CF-25E - Test（KMP）

E - Test

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Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

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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

Sample Input

Input

ab bc cd

Output

4

Input

abacaba abaaba x

Output

11

思路:枚举6种排列方式，求出它这些字符串的最大重合数，总的减去最大重合数就是答案。

      如三个串是a,b,c;

  abc,acb,bac,bca,cab,cba 六种排列。找出最大重合。而这最大重合只能是末尾，所以用个KMP匹配两个串，返回主串结束时，模式串匹配的位置就是最大匹配数。

   不过要注意一点，但中间串是前串的子串时，应该用前串与后串匹配。如abc排列，b是a的子串则a与匹配，否则，b与c匹配。

#include<iostream>
#include<cstring>
using namespace std;
const int mm=100040;
int next[3][mm],l[3];
char s[3][mm];
void get(int x)///获得每个串的next
{
  int k=-1,j=0;
  next[x][0]=-1;
  while(j<l[x])
  {
    if(k==-1||s[x][k]==s[x][j])
    {
      ++k;++j;next[x][j]=k;
    }
    else k=next[x][k];
  }
}
int kmp(int x,int y)
{
  int j=0,k=0;
  while(j<l[y]&&k<l[x])
  {
    if(j==-1||s[x][k]==s[y][j])
      {++j;++k;}
    else j=next[y][j];
  }
  return j;///返回结尾的最大匹配数
}
int main()
{
  while(cin>>s[0]>>s[1]>>s[2])
  {
    l[0]=strlen(s[0]);
    l[1]=strlen(s[1]);
    l[2]=strlen(s[2]);
    for(int i=0;i<3;i++)get(i);
    int ans=-3333;
    int z,zz=0;
    ///手动枚举六种情况
    ///012
    z=kmp(0,1);
    zz=z;
    if(l[1]==z)zz+=kmp(0,2);
    else zz+=kmp(1,2);
    ans=ans>zz?ans:zz;
    ///021
    z=kmp(0,2);
    zz=z;
    if(l[2]==z)zz+=kmp(0,1);
    else zz+=kmp(2,1);
    ans=ans>zz?ans:zz;
    ///102
    z=kmp(1,0);
    zz=z;
    if(l[0]==z)
      zz+=kmp(1,2);
    else zz+=kmp(0,2);
    ans=ans>zz?ans:zz;
    ///120
    z=kmp(1,2);
    zz=z;
    if(l[2]==z)zz+=kmp(1,0);
    else zz+=kmp(2,0);
    ans=ans>zz?ans:zz;
    ///201
    z=kmp(2,0);zz=z;
    if(l[0]==z)zz+=kmp(2,1);
    else zz+=kmp(0,1);
    ans=ans>zz?ans:zz;;
    ///210
    z=kmp(2,1);zz=z;
    if(l[1]==z)zz+=kmp(2,0);
    else zz+=kmp(1,0);
    ans=ans>zz?ans:zz;
    ///这应该还可以编个函数调用枚举，但数据小就不整了
    zz=l[0]+l[1]+l[2];
    zz=zz-ans;
    cout<<zz<<"\n";
  }
}