A + B for you again HDU - 1867（最大前缀&最大后缀的公共子缀&kmp删除法）

最新推荐文章于 2025-01-08 08:30:00 发布

donggengqi9263

最新推荐文章于 2025-01-08 08:30:00 发布

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原文链接：http://www.cnblogs.com/dwj-2019/p/11375665.html

本文介绍了一种解决字符串拼接问题的算法，通过寻找两字符串间最大公共子串，实现最短且字典序最小的拼接结果。使用KMP算法进行高效匹配。

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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

题目大意：

因为

但是如果是

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxx=100100;
int p[maxx];
char a[maxx],b[maxx];
void pre(char b[])
{
    int m=strlen(b);
    int i=1,j=0;
    p[0]=-1;
    while(i<m)
    {
        if(j==-1||b[i]==b[j])
        {
            i++,j++;
            p[i]=j;
        }
        else
            j=p[j];
    }
}
int kmp(char a[],char b[])
{
    int n=strlen(a);
    int m=strlen(b);
    pre(b);
    int i=0,j=0;
    while(i<n&&j<m)
    {
        if(j==-1||a[i]==b[j])
        {
            i++,j++;
        }
        else j=p[j];
    }
    if(i==n)
        return j;
    else
        return 0;
}
int main()
{
    while(~scanf("%s%s",a,b))
    {
        int x=kmp(a,b);
        int y=kmp(b,a);
        if(x==y)
        {
            if(strcmp(a,b)>0)
            {
                printf("%s",b);
                printf("%s",a+x);
            }
            else
            {
                printf("%s",a);
                printf("%s",b+x);
            }
        }
        else if(x>y)
        {
            printf("%s",a);
            printf("%s",b+x);
        }
        else
        {
            printf("%s",b);
            printf("%s",a+y);
        }
        printf("\n");
    }
    return 0;
}