hdu1867之KMP

 

A + B for you again

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3061    Accepted Submission(s): 755

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

 

 

注意题目要求的是最短的字符串能包含所给的两个字符串,这里的包含一定是前一部分或后一部分包含,不能中间包含

 

#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<cstring>

#include<string>

#include<algorithm>

#include<map>

#include<iomanip>

#define INF 99999999

using namespace std;



const int MAX=100000+10;

char a[MAX],b[MAX];

int next[MAX];



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(){

	while(cin>>a>>b){

		int lena=strlen(a),lenb=strlen(b);

		int la=KMP(a,b,lena,lenb);

		int lb=KMP(b,a,lenb,lena);

		if(la>lb || (la == lb && strcmp(a,b)>0)){

			cout<<b;

			for(int i=la;i<lena;++i)cout<<a[i];

		}

		else{

			cout<<a;

			for(int i=lb;i<lenb;++i)cout<<b[i];

		}

		cout<<endl;

	}

	return 0;

}