hdu3374(最小表示法+KMP)

最新推荐文章于 2021-02-10 16:16:28 发布

晓风残月xj

最新推荐文章于 2021-02-10 16:16:28 发布

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分类专栏：数据结构-最小表示法数据结构-KMP 文章标签：数据结构 KMP 字符串最小表示法

本文链接：https://blog.youkuaiyun.com/xiaofengcanyuexj/article/details/10024125

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本文探讨了如何通过字符串左移生成多个字符串，并计算出原始字符串在这些字符串中的排名及出现次数。通过KMP算法找到循环节，进一步计算最小与最大表示的起始下标及其频率。

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String Problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1296 Accepted Submission(s): 585

Problem Description

Give you a string with length N, you can generate N strings by left shifts. For example let consider the string “SKYLONG”, we can generate seven strings:
String Rank
SKYLONG 1
KYLONGS 2
YLONGSK 3
LONGSKY 4
ONGSKYL 5
NGSKYLO 6
GSKYLON 7
and lexicographically first of them is GSKYLON, lexicographically last is YLONGSK, both of them appear only once.
Your task is easy, calculate the lexicographically fisrt string’s Rank (if there are multiple answers, choose the smallest one), its times, lexicographically last string’s Rank (if there are multiple answers, choose the smallest one), and its times also.

Input

Each line contains one line the string S with length N (N <= 1000000) formed by lower case letters.

Output

Output four integers separated by one space, lexicographically fisrt string’s Rank (if there are multiple answers, choose the smallest one), the string’s times in the N generated strings, lexicographically last string’s Rank (if there are multiple answers, choose the smallest one), and its times also.

Sample Input

  
   abcder
aaaaaa
ababab

Sample Output

Author

WhereIsHeroFrom

Source

HDOJ Monthly Contest – 2010.04.04

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本题要求一个给定字符串的最小、最大表示的起始下标和其对应的出现次数。

求最小、最大表示的起始下标可以直接改动最小表示法的实现算法；而其对应的出现次数可以先求得最小、最大表示的字符串，然后运用KMP算法找循环节。

最小表示法+KMP

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

const int MAXN=1000000+100;
int next[MAXN];

//KMP算法中计算next[]数组
void getNext(char *p)
{
    int j,k,len=strlen(p);
    j=0;
    k=-1;
    next[0]=-1;
    while(j<len)
    {
        if(k==-1||p[j]==p[k])
        {
            next[++j]=++k;
        }
        else k=next[k];
    }
}

//返回母串的最小子串的起始位置
int minpresent(char *str)
{
	int i,j,k,len=strlen(str);
	i=0,j=1,k=0;
	while(i<len&&j<len&&k<len)
	{
		if(str[(i+k)%len]==str[(j+k)%len])
			k++;
		else 
		{
			if(str[(i+k)%len]>str[(j+k)%len])
			i=i+k+1;
		    else 
			j=j+k+1;
			if(i==j)j++;
			k=0;
		}
	}
	return ++i<++j?i:j;
}

//返回母串的最大子串的起始位置
int maxpresent(char *str)
{
	int i,j,k,len=strlen(str);
	i=0,j=1,k=0;
	while(i<len&&j<len&&k<len)
	{
		if(str[(i+k)%len]==str[(j+k)%len])
			k++;
		else 
		{
			if(str[(i+k)%len]<str[(j+k)%len])
			i=i+k+1;
		    else 
			j=j+k+1;
			if(i==j)j++;
			k=0;
		}
	}
	return ++i<++j?i:j;
}

char str[MAXN];
char Minstr[MAXN];
char Maxstr[MAXN];

int main()
{
	int n,i,Minid,Minlen,Minans,Maxid,Maxlen,Maxans;
	while(~scanf("%s",str))
	{
		Minstr[0]=0;
		Minid=minpresent(str)-1;
		strcpy(Minstr,str+Minid);
		strncat(Minstr,str,Minid);
	//	printf("Minstr=%s\n",Minstr);
		getNext(Minstr);
		Minlen=strlen(Minstr);
		Minans=1;
		if(Minlen%(Minlen-next[Minlen])==0)
			Minans=Minlen/(Minlen-next[Minlen]);
		//printf("Minid=%d   Minans=%d\n",Minid+1,Minans);

		Maxstr[0]=0;
		Maxid=maxpresent(str)-1;
		strcpy(Maxstr,str+Maxid);
		strncat(Maxstr,str,Maxid);
	//	printf("Maxstr=%s\n",Maxstr);
		getNext(Maxstr);
		Maxlen=strlen(Maxstr);
		Maxans=1;
		if(Maxlen%(Maxlen-next[Maxlen])==0)
			Maxans=Maxlen/(Maxlen-next[Maxlen]);
	//	printf("Maxid=%d   Maxans=%d\n",Maxid+1,Maxans);

		printf("%d %d %d %d\n",Minid+1,Minans,Maxid+1,Maxans);
	}
	return 0;
}