hdu 1867 A + B for you again (kmp)

A + B for you again

                                                                                Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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
   
   
   
   

 

题意：求str1 的最长后缀与str2的最长前缀，使得str1 + str2 的长度最小，并且字典序最小，str1 和str2 可交换。

分析：求最长后缀和最长前缀时，自然会想到kmp算法。在此题中，next[i] 表示以i为结尾的字串与开头同样长度的子串匹配的长度。在求next时，我将两个字符串进行了合并，但要注意的是，合并时要在两个字符串之间加上一个字符，不然在合并后求next时会出错（例如abcbcbca bcbcbc）。求出两次合并后最后一个字符的next值，即为最长前缀，输出时把这部分去掉就可以了。具体实现请参考代码。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 1e5 + 10;
char str1[N], str2[N], str11[N*2], str22[N*2], s[N*2];
int next[N*2];
char* Merge(char *s1, char *s2)
{
    int i, j;
    int len1 = strlen(s1);
    int len2 = strlen(s2);
    strcpy(s, s1);
    s[len1] = '.';
    for(i = len1 + 1, j = 0; j < len2; i++, j++)
        s[i] = s2[j];
    s[i] = '\0';
    return s;
}

int get_next(char *a)
{
    int i, j = 0;
    int len = strlen(a);
    for(i = len - 1; i >= 0; i--)
        a[i+1] = a[i];
    next[1] = 0;
    for(i = 2; i <= len; i++)
    {
        while(j > 0 && a[i] != a[j+1])
            j = next[j];
        if(a[i] == a[j+1])
              j++;
        next[i] = j;
    }
    return next[len];
}

int main()
{
    int i;
    while(~scanf("%s%s",str1, str2))
    {
        strcpy(str11, Merge(str1, str2));
        int p = get_next(str11);
        strcpy(str22, Merge(str2, str1));
        int q = get_next(str22);
        if(p > q) //sdfg  asdf
            printf("%s%s",str2, str1 + p);
        else if(p < q) //asdf sdfg
            printf("%s%s",str1,str2+q);
        else
        {
            if(strcmp(str11, str22) < 0)
                printf("%s%s",str1,str2+p);
            else if(strcmp(str11, str22) > 0)
                printf("%s%s",str2,str1+p);
            else
                printf("%s",str1);
        }
        printf("\n");
    }
    return 0;
}

下面的代码是对上面的优化，节省了一部分内存。

#include<stdio.h>
#include<string.h>
const int N = 1e5 + 10;
char str1[N], str2[N], s[N*2];
int next[N*2];

void get_next(char *a)
{
    int i, j = 0;
    int len = strlen(a+1);
    next[1] = 0;
    for(i = 2; i <= len; i++)
    {
        while(j > 0 && a[i] != a[j+1])
            j = next[j];
        if(a[i] == a[j+1])
              j++;
        next[i] = j;
    }
}

int get_Max_len(char *s1, char *s2)
{
    int i, j;
    int len1 = strlen(s1 + 1);
    int len2 = strlen(s2 + 1);
    int lens = len1 + len2 + 1;
    for(i = 1; i <= len1; i++)
        s[i] = s1[i];
    s[len1 + 1] = '.';
    for(i = 1, j = len1 + 2; i <= len2; i++, j++)
        s[j] = s2[i];
    s[j] = '\0';
    get_next(s);
    return next[lens];
}
int main()
{
    int i;
    while(~scanf("%s%s",str1 + 1, str2 + 1))
    {
        int p = get_Max_len(str1, str2);
        int q = get_Max_len(str2, str1);
        //printf("%d %d\n",p,q);
        if(p > q) //sdfg  asdf
            printf("%s%s",str2+1,str1+p+1);
        else if(p < q) //asdf sdfg
            printf("%s%s",str1+1, str2+q+1);
        else
        {
            if(strcmp(str1+1, str2+1) < 0)
                printf("%s%s",str1+1,str2+1+p);
            else if(strcmp(str1+1, str2+1) > 0)
                printf("%s%s",str2+1,str1+p+1);
            else
                printf("%s",str1+1);
        }
        printf("\n");
    }
    return 0;
}

有一点我没搞明白，最后输出时，单个字符的输出超时，输出字符串就AC了。