HDU 2594 next[len]应用

最新推荐文章于 2024-07-20 14:45:00 发布

ClownFeilong

最新推荐文章于 2024-07-20 14:45:00 发布

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分类专栏：竞赛算法文章标签： kmp

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本文介绍了一种用于发现字符串间最长前缀匹配的算法，该算法由《辛普森一家》中的荷马提出，旨在通过寻找两个名字之间的共同部分来揭示隐藏的才能。

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J - Simpsons’ Hidden Talents

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

应用next[len]，注意一个最大长度min(lena,lenb)。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#include<map>
#include<string>

using namespace std;

const int maxlen = 1000100;

int n;
int m;
int arrnext[maxlen+1];
int res[maxlen + 1];

char mai[maxlen];
char sub[maxlen];

void getnext()
{
	int i = 0;
	int j = -1;
	arrnext[0] = -1;
	int len = strlen(sub);
	while (i <= len)
	{
		if (j == -1 || sub[i] == sub[j])
		{
			j++;
			i++;
			arrnext[i] = j;
			//if (i-arrnext[i] != i&&i%(i-arrnext[i]) == 0)
			//{
			//	printf("%d %d\n", i, i / (i - arrnext[i]));
			//}
		}
		else
			j = arrnext[j];
	}
	//printf("%d\n",len%(len - arrnext[len])==0?len/(len - arrnext[len]):1);
}

void kmp()
{
	getnext();
	int ans = 0;
	int i = 0;
	int j = 0;
	int lenmai = strlen(mai);
	int lensub = strlen(sub);

	while (i < lenmai)
	{
		while (i < lenmai&&j < lensub)
		{
			if (j == -1 || mai[i] == sub[j])
			{
				i++;
				j++;
			}
			else
				j = arrnext[j];
		}

		if (j == lensub)
		{
			j = 0;
			ans++;
		}
	}//while

	printf("%d\n", ans);
}


int t;

int main()
{

	//scanf("%d", &t);
	int key = 0;
	while (scanf("%s", sub)!=EOF)
	{
		int lena = strlen(sub);
		scanf("%s", sub + strlen(sub));
		//printf("%s\n", mai);
		//continue;
		//int x;
		//scanf("%d", &x);

		//if (x == 0)
		//	break;
		key++;
		//scanf("%s",mai);
		//scanf("%s", sub);
		//if (sub[0] == '.')
			//break;

		//printf("Test case #%d\n", key);
		//kmp();
		getnext();
		//printf("\n");
		int lensub = strlen(sub);
		int lenb = lensub - lena;
		//int ans = lensub - arrnext[lensub];
		//printf("%d\n", lensub%ans == 0 ? (lensub == ans ? ans : 0) : ans - (lensub%ans));

		int ans = arrnext[lensub];
		int ansmax = min(lena, lenb);
		if (ans > ansmax)
			ans = ansmax;

		if (ans != 0)
		{
			for (int i = 0;i < ans;i++)
				printf("%c", sub[i]);
			printf(" ");
		}
		printf("%d\n", ans);
	}

	return 0;
}