HDU1867:A + B for you again【kmp】

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
需要匹配两次，找出最大匹配
#include <stdio.h>
#include <string.h>

int next[100005];

void getnext(char str[])
{
int i = 1,j = 0;
int len = strlen(str);
next [0] = -1;
while(i < len)
{
if(j == -1 || str[i] == str[j])
{
i++;
j++;
next[i] = j;
}
else
j = next[j];
}
}

int kmp(char str1[],char str2[])
{
int i= 0,j = 0;
int len1 = strlen(str1),len2 = strlen(str2);
getnext(str2);
while(i<len1 && j<len2)
{
if(j == -1 || str1[i] == str2[j])
{
i++;
j++;
}
else
j = next[j];
}
if(i == len1)
return j;
return 0;
}

int main()
{
int x,y;
char str1[100005],str2[100005];
while(scanf("%s%s",str1,str2)!=EOF)
{
x = kmp(str1,str2);
y = kmp(str2,str1);
if(x == y)  //x==y说明两者有一个是另一个的子串
{
if(strcmp(str1,str2)>0)  //str2是子串
{
printf("%s",str2);
printf("%s\n",str1+x);
}
else
{  //str1是子串
printf("%s",str1);
printf("%s\n",str2+x);
}
}
else if(x>y)
{
printf("%s",str1);
printf("%s\n",str2+x);
}
else
{
printf("%s",str2);
printf("%s\n",str1+y);
}
}

return 0;
}