HDU 2594 Simpsons’ Hidden Talents KMP

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原文链接：http://www.cnblogs.com/hebozi/archive/2012/08/06/2624581.html

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#java #php

本文探讨了家庭成员Homer如何通过寻找名字中隐藏的政治才能来揭示个人潜力的故事。通过计算名字中最长前缀作为后缀的长度，揭示了深藏不露的才能。文章包含了一个解决此类问题的代码实例，涉及字符串操作和算法应用。

Simpsons’ Hidden Talents

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 833 Accepted Submission(s): 280

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
  

代码：

#include<stdio.h>
#include<string.h>
#define N 110000
#define MIN(a,b) (a<b?a:b)
char s[N],ss[N];
int next[N];
int p;

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

int main()
{
   while(scanf("%s%s",s,ss)!=-1)
   {
     int n=MIN(strlen(s),strlen(ss));
strcat(s,ss);
getnext();
int i=next[p];
while(i>n&&i>0) //此处不懂；
i=next[i];
s[i]=0;
if(i!=0)
printf("%s %d\n",s,i);
else
printf("%d\n",i);
   }
   return 0;
}