HDU A + B for you again （KMP）

最新推荐文章于 2025-01-15 08:45:00 发布

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最新推荐文章于 2025-01-15 08:45:00 发布

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文章标签： java

本文介绍了一个关于字符串处理的问题，特别是如何通过KMP算法找到两个字符串的公共子串，并据此进行字符串的连接操作，以实现最短和字典序最小的目标。

A + B for you again

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

Total Submission(s) : 30 Accepted Submission(s) : 9

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

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int maxn=100010;

int next[maxn];
char s1[maxn],s2[maxn];

void getnext(char *t){
    int len=strlen(t);
    int i=0,j=-1;
    next[0]=-1;
    while(i<len){
        if(j==-1 || t[i]==t[j]){
            i++;j++;
            next[i]=j;
        }else
            j=next[j];
    }
}

int KMP(char *s,char *t){
    int len1=strlen(s),len2=strlen(t);
    int i=0,j=0;
    getnext(t);
    while(i<len1 && j<len2){
        if(j==-1 || s[i]==t[j]){
            i++;j++;
        }else
            j=next[j];
    }
    if(i==len1)
        return j;
    return 0;
}

int main(){

    //freopen("input.txt","r",stdin);

    while(~scanf("%s%s",s1,s2)){
        int x=KMP(s1,s2);
        int y=KMP(s2,s1);
        if(x==y){
            if(strcmp(s1,s2)<0)
                printf("%s%s\n",s1,s2+x);
            else
                printf("%s%s\n",s2,s1+x);
        }else if(x>y)
            printf("%s%s\n",s1,s2+x);
        else
            printf("%s%s\n",s2,s1+y);
    }
    return 0;
}