HDU 1867 A + B for you again 【KMP】

题目链接

题目：

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 sdfgasdf ghjk

 

Sample Output

asdfgasdfghjk

思路：KMP

WA：题目意思比较隐晦，要判断a+b和b+a的情况如果两个长度相同按字典序输出

AC代码：

#include <iostream>#include <stdio.h>#include <cstring>#define N 100010using namespace std;char a[N],b[N];int nx[N];void getnext(char x[], int m){    nx[0] = 0;    int j = 0;    for(int i = 1; i < m; i++)    {        while(x[i] != x[j] && j)            j = nx[j-1];        if(x[i] == x[j])            j++;        nx[i] = j;    }}int KMP(char x[], char y[], int n, int m){    int j = 0;    for(int i = 0; i < n; i++)    {        while(x[i] != y[j] && j)            j = nx[j-1];        if(x[i] == y[j])            j++;        if(j == m && i != n-1)            j = nx[j-1];    }    return j;}int main(){    while(~scanf(" %s %s", a, b))    {        int n = strlen(a);        int m = strlen(b);        getnext(b,m);        int k = KMP(a,b,n,m);        getnext(a,n);        int kk = KMP(b,a,m,n);        if(k == kk)        {            if(strcmp(a,b) < 0)                printf("%s%s\n",a,b+k);            else                printf("%s%s\n",b,a+k);        }        else if(k > kk)            printf("%s%s\n", a, b+k);        else            printf("%s%s\n", b, a+kk);    }    return 0;}