hdu2594 Simpsons’ Hidden Talents（KMP）

最新推荐文章于 2019-10-20 19:09:49 发布

Renl1000

最新推荐文章于 2019-10-20 19:09:49 发布

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本文介绍了一个有趣的字符串匹配问题，即寻找两个字符串间最长的共同前后缀，并通过KMP算法实现了解决方案。

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Simpsons’ Hidden Talents

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7787 Accepted Submission(s): 2755

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

题意：输入s1，s2，求既是a的前缀又是b的后缀的最长子串和长度。

题解：kmp算法，用s2当主串，匹配算法循环结束后，s2的后j项和s1的前j项是匹配的，就直接输出s1的前j项就行了。

代码：

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define MAX 50005
char s1[MAX], s2[MAX];
int nextt[MAX];
int len1, len2, ans;

void getnext(char s1[])
{
	int i = 0;
	int j = -1;
	nextt[0] = -1;
	while(i < len1)
	{
		if(j == -1 || s1[i] == s1[j])
		{
			i++;
			j++;
			if(s1[i]!=s1[j])
				nextt[i] = j;
			else
				nextt[i] = nextt[j];
		}
		else
			j = nextt[j];
	}
}

int kmp(char s1[], char s2[])
{
	int i= 0, j = 0;
	getnext(s1);
	while(i < len2)
	{
		if(j==-1||s2[i]==s1[j])
		{
			i++;
			j++;
		}
		else
			j = nextt[j];
	}
	return j;
}

int main()
{
	while(gets(s1))
	{
		gets(s2);
		len1 = strlen(s1);
		len2 = strlen(s2);
		ans = kmp(s1, s2);
		if(ans)
		{
			for(int k = 0; k < ans; k++)
				cout << s1[k];
			cout << " " << ans << endl;
		}
		else
			cout << ans << endl;
	}
	return 0;
}