hdu 1867 A+B again for you

A + B for you again

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9291    Accepted Submission(s): 2274

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

 

简单的应用kmp以及next数组。。

#include <cstdio>
#include <iostream>
#include <string.h>
#include <string> 
#include <map>
#include <queue>
#include <deque>
#include <vector>
#include <set>
#include <algorithm>
#include <math.h>
#include <cmath>
#include <stack>
#include <iomanip>
#define mem0(s1) memset(s1,0,sizeof(s1))
#define meminf(s1) memset(s1,0x3f,sizeof(s1))
#define ll long long
using namespace std;
int nex[1000005],l1,l2;
void getn(int n,char c[])
{
    int i=0,j=-1;
    nex[0]=-1;
    while(i<n)
    {
        if(j==-1||c[i]==c[j])
        {
            i++;j++;nex[i]=j;
        }
        else j=nex[j];
    }
    return;
}
int kmp(char s1[],char s2[])
{
    int i,j=0,n,m;
    n=strlen(s1);
    m=strlen(s2);
    getn(m,s2);
    for(i=0;i<n;i++)
    {
        while(j&&s1[i]!=s2[j])j=nex[j];
        if(s1[i]==s2[j])j++;
    }
    return j;
}
int main()
{
    char s1[100005],s2[100005];
    wshile(~scanf("%s%s",s1,s2))
    {
        l1=strlen(s1);l2=strlen(s2);
        int h=kmp(s1,s2);
        int hh=kmp(s2,s1);
        if(h>hh||(h==hh&&strcmp(s1,s2)<0)) printf("%s%s\n",s1,s2+h);
        else printf("%s%s\n",s2,s1+hh);
    }
    return 0;
}