【HDU】2594Simpsons’ Hidden Talents（加上一段扰乱字符+next数组）-优快云博客

本文介绍了一种使用KMP算法解决字符串前缀和后缀匹配问题的方法，通过实例讲解如何找到两个字符串间最长的共同前缀和后缀，重点在于KMP算法的next数组应用。

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15853 Accepted Submission(s): 5403

Problem Description

Homer: Marge, I just figured out a way to discover some of the talents we weren’t aware we had.
Marge: Yeah, what is it?
Homer: Take me for example. I want to find out if I have a talent in politics, OK?
Marge: OK.
Homer: So I take some politician’s name, say Clinton, and try to find the length of the longest prefix
in Clinton’s name that is a suffix in my name. That’s how close I am to being a politician like Clinton
Marge: Why on earth choose the longest prefix that is a suffix???
Homer: Well, our talents are deeply hidden within ourselves, Marge.
Marge: So how close are you?
Homer: 0!
Marge: I’m not surprised.
Homer: But you know, you must have some real math talent hidden deep in you.
Marge: How come?
Homer: Riemann and Marjorie gives 3!!!
Marge: Who the heck is Riemann?
Homer: Never mind.
Write a program that, when given strings s1 and s2, finds the longest prefix of s1 that is a suffix of s2.

Input

Input consists of two lines. The first line contains s1 and the second line contains s2. You may assume all letters are in lowercase.

Output

Output consists of a single line that contains the longest string that is a prefix of s1 and a suffix of s2, followed by the length of that prefix. If the longest such string is the empty string, then the output should be 0.
The lengths of s1 and s2 will be at most 50000.

Sample Input

clinton homer riemann marjorie

Sample Output

0 rie 3

Source

HDU 2010-05 Programming Contest

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题目大意：现在给你两个字符串然后问你这两个字符串的前缀和后缀的相同的数目是多少

思路：其实看到这个前缀后缀就可以想到我们的KMP算法中的next数组了，然后如果直接相加的话是错误的，因为可能你加上的后面的数组后变成了一个前缀后缀特别大的字符串了，为了防止前面字符串的尾巴和下面的有一些别的‘反应’，我们知道加上那个干扰字符串就好了，我的话是加上了两个，貌似这个题目加上一个也行，但是总觉得怪怪的。。。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<string>
using namespace std;
const int maxn=50000*2+5;
int nxt[maxn];
void getnext(char p[])
{
    nxt[0]=-1;
    int j=0,k=-1;
    int len=strlen(p);
    while(j<len)
    {
        if(k==-1||p[j]==p[k])
            nxt[++j]=++k;
        else
            k=nxt[k];
    }
}
char s1[maxn],s2[maxn];
int main()
{
    while(~scanf("%s %s",&s1,&s2))
    {
        int m=strlen(s1);
        char str[]={'#','$'};
        strcat(s1,str);
        //cout<<s1<<endl;
        strcat(s1,s2);
        getnext(s1);
        int len=strlen(s1);
        if(nxt[len]==0)
            cout<<0<<endl;
        else
        {
            for(int i=0;i<nxt[len];i++)
                cout<<s1[i];
            cout<<" "<<nxt[len]<<endl;
        }
    }
}