hdu 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：

给你两个字符串a ,b，在输出要求下输出a+b或者b+a。先说一下这个‘+“法的规定。若a的某一最大后缀和b的某一最大前缀相等，则可合并。
优先保证字符串最短，如果相同输出字典序小的那一个

题解：

KMP算法的应用。自己wrong answer了十几次，参考了大佬的代码
地址：http://blog.csdn.net/libin56842/article/details/8496648

代码：

#include <stdio.h>#include <string.h>#include <iostream>using namespace std;int ne[100005];void getnext(char str[]){    int i = 1,j = 0;    int len = strlen(str);    ne [0] = -1;    while(i < len)    {        if(j == -1 || str[i] == str[j])        {            i++;            j++;            ne[i] = j;        }        else            j = ne[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 = ne[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)        {            if(strcmp(str1,str2)>0)            {                printf("%s",str2);                printf("%s\n",str1+x);            }            else            {                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;}