hdu 2594 Simpsons’ Hidden Talents

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)

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

题目大意：输出最大的第一个字符串的前缀=第二个字符串的后缀长度

还记得next 数组的含义吗：next[i]=j表示i的后缀字符串=前缀字符串的最大长度

但这里是2个字符串，所以把他们接起来就好了。

注意答案不能直接输出next[len1+len2]，因为有可能他超过了第2个字符串的长度

所以要从后往前找到第一个next[i]既小于len1，又小于len2的next[i] 

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
char s1[50001],s2[50001];
int len1,len2,f[100001];
void getnext()
{
    strcat(s1,s2);
    for(int i=1;i<len1+len2;i++)
    {
        int j=f[i];
        while(j&&s1[j]!=s1[i]) j=f[j];
        f[i+1]= s1[i]==s1[j] ? j+1:0;
    }
}
int main()
{
    while(cin>>s1>>s2)
    {
        len1=strlen(s1);
        len2=strlen(s2);
        getnext();
        int d=len1+len2;
        while(d>len1||d>len2) d=f[d];
        for(int i=0;i<d;i++) printf("%c",s1[i]);
        if(d) printf(" ");
        printf("%d\n",d);
    }
}