A + B for you again(KMP)

男娘i 2022-08-02 13:42 8阅读 0赞

# A + B for you again #

**Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 4940    Accepted Submission(s): 1248**

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

Author

Wang Ye

Source

[2008杭电集训队选拔赛——热身赛][2008]

#include<iostream>
    #include<cstdio>
    #include<cstring>
    using namespace std;
    const int MAXN = 100005,MAXM = 100005;
    char A[MAXN],B[MAXM];
    int Next[MAXM];
    void get_next(char *a,int len){
        int i=-1,j=0;
        Next[0]=-1;
        while(j<len){
            if(i == -1 || a[i] == a[j]){
                if(a[++i] == a[++j])Next[j]=Next[i];
                else Next[j]=i;
            }else i=Next[i];
        }
    }
    
    int KMP(char *a,char *b,int lena,int lenb){
        get_next(a,lena);
        int i=0,j=0;
        while(i<lena && j<lenb){
            if(i == -1 || a[i] == b[j])++i,++j;
            else i=Next[i];
        }
        if(i<lena || (i == lena && j == lenb))return i;//a不能是b中间部分的字串
        return 0;
    }
    
    int main()
    {
        int len1,len2,p1,p2,i;
        while(cin>>A>>B)
        {
            //cout<<T<<" "<<P<<endl;
            len1=strlen(A);
            len2=strlen(B);
            p1=KMP(A,B,len1,len2);//B中找A，找到的最大长度
            p2=KMP(B,A,len2,len1);//A中找B,找到的最大长度
            /*
            比如：asdf sdfg
            在B中根本找不到A的子串（必须从头开始）
            而在A中却能找到B的子串最大长度为3
            所以输出 0 3
            cout<<p1<<" "<<p2<<endl;*/
            if(p1==p2)
            {
                if(strcmp(A,B)<0)
                {
                    cout<<A;
                    cout<<B+p2<<endl;
                }
                else
                {
                    cout<<B;
                    cout<<A+p1<<endl;
                }
            }
            else
            {
                if(p1>p2)//说明在B中找到了A的子串
                {
                    cout<<B;
                    cout<<A+p1<<endl;
                }
                else//在A中找到了B的子串
                {
                    cout<<A;
                    cout<<B+p2<<endl;
                }
            }
        }
    }

[2008]: http://acm.hdu.edu.cn/search.php?field=problem&key=2008%BA%BC%B5%E7%BC%AF%D1%B5%B6%D3%D1%A1%B0%CE%C8%FC%A1%AA%A1%AA%C8%C8%C9%ED%C8%FC&source=1&searchmode=source