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

注意題目要求的是最短的字符串能包含所給的兩個字符串,這裡的包含一定是前一部分或後一部分包含,不能中間包含

#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中間部分的字串    
    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;  
}   

#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中間部分的字串 
	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;
}