CodeForces 25E Test KMP

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

4

Input

abacaba
abaaba
x

Output

11

利用kmp匹配算法，在kmp匹配算法稍加修改就ok，因为这里可以只匹配一部分前缀，不用完全匹配。

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;

char str[1100000];
char s0[1100000],s1[1100000],s2[1100000];
int next[1100000];
void getnextval(char *p)//求一个字符串的next数组，结果保存在全局变量next[]中。
{
     int plen=strlen(p);
     next[0]=-1;
     int k=-1;
     int j=0;
     while (j<plen-1)
     {
           if (k==-1||p[k]==p[j])
           {
                  j++;
                  k++;
                  if (p[j]!=p[k])
                  next[j]=k;
                  else
                  next[j]=next[k];
           }
           else
           {
               k=next[k];
           }
     }
}

int kmp(char *s,char *p)//以s为匹配串，p为模式串进行匹配，返回匹配后p剩下的字符个数。
{
    int l;
    int i=0;
    int j=0;
    int slen=strlen(s);
    int plen=strlen(p);
    getnextval(p);
    while (i<slen&&j<plen)
    {
          if (j==-1||s[i]==p[j])
          {
               i++;
               j++;
          }
          else
          j=next[j];
    }
    if (j==plen)//完全匹配的情况，一个字符也不剩，放回0。
    l=0;
    else if (i==slen)  //匹配一部分，也就是p的一个前缀和s的一个后缀完全匹配。
    l=plen-j;

    else
    l=plen-(slen-i)-1;

    return l;
}

int cat(char *a,char *b,char *c)   //假设a串在最前，b串在中间，c串在最后至少需要多长。
{
    str[0]='\0';
    int la=strlen(a);
    int lb=strlen(b);
    int lc=strlen(c);
    int len=kmp(a,b);
    strcat(str,a);
    strcat(str,b+lb-len);
    len=kmp(str,c);
    return strlen(str)+len;
}

int main()
{
    while (scanf("%s%s%s",s0,s1,s2)!=EOF)
    {
          int minn;
          int l;
          //一下分6种情况讨论。
          minn=cat(s0,s1,s2);

          l=cat(s0,s2,s1);
          if (l<minn)
          minn=l;

          l=cat(s1,s2,s0);
          if (l<minn)
          minn=l;

          l=cat(s1,s0,s2);
          if (l<minn)
          minn=l;

          l=cat(s2,s0,s1);
          if (l<minn)
          minn=l;

          l=cat(s2,s1,s0);
          if (l<minn)
          minn=l;

          cout <<minn<<endl;
    }
    return 0;
}

　　对于kmp的学习推荐博客：http://blog.csdn.net/v_july_v/article/details/7041827，讲的很好。