HDU 2594 Simpsons’ Hidden Talents (KMP——前缀后缀最长公共串)

Simpsons’ Hidden Talents

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4521    Accepted Submission(s): 1639

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：这题就是求第一个字符串的前缀和第二个字符串的后缀的最长公共字串，不存在输出0

分析：这题和POJ 2752差不多，都是求前缀和后缀的公共串，这题就是把两个字符串连一起就行了，要注意，如果nex[n]超过一个串的长度的话，输出短的字符串。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<string>
#include<iostream>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>
#include<vector>
#include<map>
#include<stdlib.h>
#include<algorithm>
#define LL __int64
#define FIN freopen("in.txt","r",stdin)
using namespace std;
const int MAXN=50000+5;
char str[MAXN*2],b[MAXN];
int nex[MAXN*2];
int n;

void getnext()
{
    int j=0,k=-1;
    nex[0]=-1;
    while(j<n)
    {
        if(k==-1 || str[j]==str[k])
            nex[++j]=++k;
        else
            k=nex[k];
    }
}

int main()
{
    //FIN;
    while(scanf("%s %s",str,b)!=EOF)
    {
        int len1=strlen(str),len2=strlen(b);

        strcat(str,b);
        n=strlen(str);
        getnext();
        while(nex[k]>len1 || nex[k]>len2)
            k=nex[k];

        if(nex[n]==0) printf("0
");
        else
        {
            for(int i=0;i<len;i++)
                printf("%c",str[i]);
            printf(" %d
",len);
        }
    }
    return 0;
}