hdu 1867：A + B for you again（kmp）

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

题意分析：

两个字符串a,b,如果a的末尾和b的开头相同或者b的末尾和a的开头相同，相同部分只输出一次。输出可以去掉字符最多的那一个，如果相同输出字典序最小的。

解题思路：

假设a在前面，求相同的字符有多少，用b做子串， 匹配之后，如果匹配到了a的末尾，那么就是末尾与b有相同的，而此时b的匹配位置j就是相同长度。

b在前面时道理是一样的，分别求出两个重复部分比较输出。

#include <stdio.h>
#include <string.h>
#define N 100020
char str1[N], str2[N];
int next[N];
void get_next(char *b, int lenb)
{
	memset(next, 0, sizeof(next));
	int i=1, j=0;
	next[0]=-1;
	while(i<lenb)
	{
		if(j==-1 || b[i]==b[j])
		{
			i++;
			j++;
			next[i]=j;
		}
		else j=next[j];
	}
}
int kmp(char *a, char *b, int lena, int lenb)
{
	get_next(b, lenb);
	int i=0, j=0;
	while(i<lena && j<lenb)
	{
		if(j==-1 || a[i]==b[j])
		{
			i++;
			j++;
		}
		else j=next[j];
	}
	if(i==lena)
		return j;
	return 0;
}
int main()
{
	int lena, lenb, k1, k2;
	while(scanf("%s%s", str1, str2)!=EOF)
	{
		lena=strlen(str1);
		lenb=strlen(str2);
		k1=kmp(str1, str2, lena, lenb);
		k2=kmp(str2, str1, lenb, lena);
		if(k1>k2)
			printf("%s%s
", str1, str2+k1);
		else if(k1<k2)
			printf("%s%s
", str2, str1+k2);
		else{
			if(strcmp(str1, str2)<0)
				printf("%s%s
", str1, str2+k1);
			else
				printf("%s%s
", str2, str1+k1);
		}
	}
	return 0;
}