hdu 1867 A + B for you again

A + B for you again

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4661    Accepted Submission(s): 1198

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

题目大意：两个字符串合并成一个，串1的后缀和串2前缀相同部分只出现一次，不固定串1串2，要求合并后串长最小，如果有等串长的两种情况出现，取字典序小的。

解题思路：利用kmp算法的next数组，先将两个字符串守卫相连，若有相同的前缀后缀，减去后缀，交换两个字符串输出，细节见代码。

代码如下：

#include <cstdio>
#include <cstring>
const int maxn=1000010;
char a[maxn],b[maxn];
char s[maxn];
char s1[maxn];
int next[maxn];
int len;
void get_next()
{
    int i=0,j=-1;
    next[0]=-1;
    while(i<len)
    {
        if(j==-1||s[i]==s[j])
        {
            i++;
            j++;
            next[i]=j;
        }
        else
            j=next[j];
    }
}

int main()
{
    int i,cnt1,cnt2,len1,len2,cur,p;
    while(scanf("%s%s",a,b)!=EOF)
    {
        len1=strlen(a);
        len2=strlen(b);
		memset(next,0,sizeof(next));
        cur=0;
        for(i=0;i<len1;i++)   //将两个字符串首尾相连成串s
            s[cur++]=a[i];
        for(i=0;i<len2;i++)
            s[cur++]=b[i];
        len=len1+len2;
        get_next();
        cnt1=next[len];    //求得最长的相同前缀后缀的长度
		if(cnt1>len2) cnt1=len1;     //如果长度大于串2长度，说明整个串1出现在串2 中，因此将减去串1的长度，得到目标串长。

        memset(next,0,sizeof(next));
        cur=0;
        for(i=0;i<len2;i++)  //同理交换串1串2的顺序，目的是找出最大的cnt减去，得到最小串长
            s[cur++]=b[i];
        for(i=0;i<len1;i++)
            s[cur++]=a[i];
        get_next();
        cnt2=next[len];
		if(cnt2>len1) cnt2=len2;
        if(cnt1>cnt2)
        {
            for(i=0;i<len2-cnt1;i++)
                printf("%c",b[i]);
            for(i=0;i<len1;i++)
                printf("%c",a[i]);
            printf("\n");
        }
        else if(cnt1<cnt2)
        {
            for(i=0;i<len1-cnt2;i++)
                printf("%c",a[i]);
            for(i=0;i<len2;i++)
                printf("%c",b[i]);
            printf("\n");
        }
        else    //  判断是否满足字典序条件
        {
            p=0;
            for(i=0;i<len1-cnt2;i++)
                s[i]=a[i];
            cur=len1-cnt2;
            for(i=0;i<len2;i++)
                s[cur++]=b[i];
            
            for(i=0;i<len2-cnt1;i++)
                s1[i]=b[i];
			cur=len2-cnt1;
			for(i=0;i<len1;i++)
				s1[cur++]=a[i];
            for(i=0;i<len;i++)
            {
                if(s[i]>s1[i])break;
                if(s[i]<s1[i])
                {
                    p=1;break;
                }
            }
            if(p)
            {
                for(i=0;i<len-cnt1;i++)
                    printf("%c",s[i]);
                printf("\n");
            }
            else
            {
                for(i=0;i<len-cnt1;i++)
                    printf("%c",s1[i]);
                printf("\n");
            }
        }
    }
    return 0;
}