HDU 1867 A + B for you again (kmp的运用)

A + B for you again

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9063    Accepted Submission(s): 2194

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

 

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题目意思很简单，不过有点坑的地方在于最后输出的字符串要是最短的并且字典序最小的，

用kmp去找两个字符串相互重叠（ａ的后缀和b的前缀或者ｂ的前缀和a的后缀）的部分长度是多少,如果a和ｂ 相互重叠的部分一样长,就比较a和ｂ谁的字典序小，然后输出小的那个就好了．

#include <iostream>
#include<string>
#include<cstring>
using namespace std;
string a,b;
int nexts[100005];
void getnext(string b)
{
    memset(nexts,0,sizeof(nexts));
    nexts[0]=-1;
    int k=-1;
    int i=0;
    int len=b.size();
    while(i<len)
    {
        if(k==-1||b[i]==b[k])
        {
            ++i,++k;
            nexts[i]=k;
        }
        else
            k=nexts[k];
    }
}

int kmp(string a,string b)
{
    getnext(b);
    int cnt=0;
    int j=0;
    int i=0;
    int lena=a.size();
    int lenb=b.size();
    while(i<lena)
    {
        if(j==-1||a[i]==b[j])
            ++i,++j;
        else
            j=nexts[j];
    }
    return j; // 返回匹配到的最大的长度
}


int main()
{

    while(cin>>a>>b)
    {
        int len1=kmp(a,b);
        int len2=kmp(b,a);
        if(len1==len2){
            if(a<b)
                cout<<a<<b.c_str()+len1<<endl;　//巧妙的一点：string 转char＊　类型后,直接加上可重叠的长度即可输出重叠后的字符
            else
                cout<<b<<a.c_str()+len2<<endl;
        }
        else
        if(len1>len2){            //最后的结果长度尽可能短.
            cout<<a<<b.c_str()+len1<<endl;
        }
        else
            cout<<b<<a.c_str()+len2<<endl;
    }
    return 0;
}