HDU——T 2594 Simpsons’ Hidden Talents

http://acm.hdu.edu.cn/showproblem.php?pid=2594

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9919 Accepted Submission(s): 3418

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

题意：求出第一个串的最长前缀同时是后一个串的最长后缀

而kmp算法中next数组其实就是查找某串中每一位前面的子串的前后缀有多少位匹配

所以可以将两个串合并，求出next[L(L=l1+l2)]，最后保证L<=l2&&L<=l1

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cstdio>
 5 
 6 using namespace std;
 7 
 8 const int N(50000+5);
 9 char s1[N],s2[N],s3[N<<1];
10 int p[N<<1],l1,l2,l3;
11 
12 inline void Get_next()
13 {
14     for(int i=2,j=0;i<=l1+l2;i++)
15     {
16         for(;j>0&&s3[j+1]!=s3[i];) j=p[j];
17         if(s3[j+1]==s3[i]) j++;
18         p[i]=j;
19     }
20 }
21 
22 int main()
23 {
24     for(;cin>>s1+1>>s2+1;)
25     {
26         l1=strlen(s1+1); l2=strlen(s2+1);
27         for(int i=1;i<=l1;i++) s3[i]=s1[i];
28         for(int i=1;i<=l2;i++) s3[i+l1]=s2[i];
29         Get_next(); int l=l1+l2;
30         for(;l>l1||l>l2;) l=p[l];
31         for(int i=1;i<=l;i++) printf("%c",s1[i]);
32         if(l) printf(" %d
",l);
33         else puts("0");
34     }
35     return 0;
36 }

cin很灵性、、、

——每当你想要放弃的时候，就想想是为了什么才一路坚持到现在。