【KMP】 hdu1867 A + B for you again

A + B for you again

http://acm.hdu.edu.cn/showproblem.php?pid=1867

Problem Description

Generally speaking, there are a lot of problems about strings processing. Now you encounter another such problem. If you get two strings, such as “asdf” and “sdfg”, the result of the addition between them is “asdfg”, for “sdf” is the tail substring of “asdf” and the head substring of the “sdfg” . However, the result comes as “asdfghjk”, when you have to add “asdf” and “ghjk” and guarantee the shortest string first, then the minimum lexicographic second, the same rules for other additions.

 

Input

For each case, there are two strings (the chars selected just form ‘a’ to ‘z’) for you, and each length of theirs won’t exceed 10^5 and won’t be empty.

 

Output

Print the ultimate string by the book.

 

Sample Input

   
   
   
   

    
    
    
    asdf sdfg
asdf ghjk
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    asdfg
asdfghjk
   
   
   
   

#include<cstdio>
#include<cstring>
using namespace std;
#define MAX 100005
int next[MAX];
void get_next(char *t)
{
    int i=0,j=-1;
    next[0]=-1;
    for(; t[i];)
        if(j==-1||t[i]==t[j])
        {
            ++i;
            ++j;
            next[i]=j;
        }
        else  j=next[j];
}
int kmp(char *s,char *t)
{
    int i=0,j=0;
    get_next(t);
    int n=strlen(s),m=strlen(t);
    for(; i<n&&j<m;)
        if(j==-1||s[i]==t[j])
            ++i,++j;
        else
            j=next[j];
    if(i>=n)
        return j;
    return 0;
}
int main()
{
    char s[MAX],t[MAX];
    for(; ~scanf("%s%s",s,t);)
    {
        int x=kmp(s,t);
        int y=kmp(t,s);
        if(x==y)
        {
            if(strcmp(s,t)>0)
            {
                printf("%s",t);
                printf("%s\n",s+x);
            }
            else
            {
                printf("%s",s);
                printf("%s\n",t+x);
            }
        }
        else if(x>y)
        {
            printf("%s",s);
            printf("%s\n",t+x);
        }
        else
        {
            printf("%s",t);
            printf("%s\n",s+y);
        }
    }
    return 0;
}