kmp（2）

A + B for you again

Time Limit : 5000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 6   Accepted Submission(s) : 2

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 sdfgasdf ghjk

 

Sample Output

asdfgasdfghjk

 

Author

Wang Ye

 

Source

2008杭电集训队选拔赛——热身赛

题意：将两个字符串中相同部分去除，前一个串的后半部分与后一个字符串的前半部分匹配。给出两个串，问最短的串是什么。首先保证的是最短，然后保证按照字典序。

思路：kmp。由于不知道哪一个匹配另一个的字符多，所以要两个都匹配然后判断，相同时在按照字典序输出。

#include<iostream>#include<stdio.h>#include<string.h>#include<memory.h>using namespace std;char a[100005],b[100005];int next[100005];void getNext(char *b,int m){    memset(next,0,sizeof(next));    for(int i=1; i<m; i++)    {        int tmp=next[i-1];        while(tmp&&b[tmp]!=b[i])            tmp=next[tmp-1];        if(b[tmp]==b[i])            next[i]=tmp+1;        else            next[i]=0;    }}int kmp(char *a,char *b,int n){    int j=0,max=-1;    for(int i=0; i<n; i++)    {        while(j>0&&a[i]!=b[j])            j=next[j-1];        if(a[i]==b[j])            j++;        else            j=0;        //if(max<j)max=j;//j表示的是到i位置已经匹配的个数    }    return j;}int main(){    int t,m,n;    while(scanf("%s %s",&a,&b)!=EOF)    {        n=strlen(a);        m=strlen(b);        getNext(b,m);        int j=kmp(a,b,n);        getNext(a,n);        int l=kmp(b,a,m);        if(j==l)        {            if(strcmp(a,b)<=0)            {                for(int i=0; i<n; i++)                    printf("%c",a[i]);                for(int i=j; i<m; i++)                    printf("%c",b[i]);                printf("\n");            }            else            {                for(int i=0; i<m; i++)                    printf("%c",b[i]);                for(int i=l; i<n; i++)                    printf("%c",a[i]);                printf("\n");            }        }        else if(j>l)        {            for(int i=0; i<n; i++)                printf("%c",a[i]);            for(int i=j; i<m; i++)                printf("%c",b[i]);            printf("\n");        }        else        {            for(int i=0; i<m; i++)                printf("%c",b[i]);            for(int i=l; i<n; i++)                printf("%c",a[i]);            printf("\n");        }    }    return 0;}/*asdf sdfgasdf ghjk*/