A + B for you again KMP

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A + B for you again

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6352    Accepted Submission(s): 1566

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

 

Recommend

lcy

 

思路：这题就是日了狗的KMP........       给定两个字符串a和b，现在要将a和b相加，a+b相加规则为找出a最长的后缀等于b的等长前缀，之后结果为a的前面部分+相等部分+b的后面部分，例如：asdf+sdfg=asdfg。相加的时候也可以b+a，输出相加后长度最小的，若存在两者长度相等，则输出字典树较小的（要用到strcmp ）。

用一个串匹配另一个串，被匹配串的最后的一位匹配值就是最长的相等部分了。然后用这种方法求得a+b和b+a，之后选择小的输出。

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

const int maxn=1e5+10;
char a[maxn],b[maxn],c[maxn];
int f[maxn];
void GetNext(char *a,int *f,int n)
{
    int i,j;
    f[0]=f[1]=0;
    for(i=1;i<n;i++)
    {
        j=f[i];
        while(j&&a[i]!=a[j])j=f[j];
        f[i+1]=a[i]==a[j]?j+1:0;
    }
}
int KMP(char *a,char *b,int *f)
{
    int i,j=0,n,m;
    n=strlen(a);
    m=strlen(b);
    GetNext(b,f,m);
    for(i=0;i<n;i++)
    {
        while(j&&a[i]!=b[j])j=f[j];
        if(a[i]==b[j])j++;
    }
    return j;
}
int main()
{
    while(scanf("%s%s",a,b)!=EOF)
    {
        int i,j,k,p,q;
        p=KMP(a,b,f);
        q=KMP(b,a,f);
        if(p>q||p==q&&strcmp(a,b)<0) printf("%s%s\n",a,b+p);
        else printf("%s%s\n",b,a+q);
    }
    return 0;
}