hdu 2594 Simpsons’ Hidden Talents-优快云博客

本文介绍了一道关于寻找两个字符串间最长公共前后缀的问题，通过KMP算法中的next数组来高效解决。提供了完整的代码实现。

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Simpsons’ Hidden Talents

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10618 Accepted Submission(s): 3710

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

kmp算法，需要非常了解next数组，充分理解其含义。

题意：s1的前缀和后缀最长，有的话，输出这个字符串，和长度，没有就输出0.

利用next数组的特性。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 50005;
char a[2*N],b[N];
int Next[2*N];
void getNext()
{
	Next[0]=-1;
	int i=0,j=-1,len=strlen(a);
	while(i<len)
	{
		if(j==-1||a[i]==a[j])
		{
			i++;
			j++;
			Next[i]=j;
		}
		else
			j=Next[j];
	}
}
int main()
{
    while(~scanf("%s%s",a,b))
    {
    	int len=strlen(a);
    	int len1=strlen(b);
    	strcat(a,b);
    	getNext();
    	int ans=Next[len+len1];
    	while(ans>len||ans>len1)
    		ans=Next[ans];
    	a[ans]=0;
    	if(ans==0)
    		printf("0\n");
    	else
    		printf("%s %d\n",a,ans);
    }
    return 0;
}

参考网址：

http://blog.youkuaiyun.com/r131303/article/details/25426207