HDU3374 String Problem —— 最小最大表示法 + 循环节

题目链接：https://vjudge.net/problem/HDU-3374

 

String Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3646    Accepted Submission(s): 1507

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

1 1 6 1 1 6 1 6 1 3 2 3

 

 

Author

WhereIsHeroFrom

 

 

Source

HDOJ Monthly Contest – 2010.04.04

 

 

Recommend

lcy

 

 

题解：

1.求最小最大表示法，直接套。

2.怎么知道最小最大表示出现了几次呢？求最小循环节。

 

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <sstream>
#include <algorithm>
using namespace std;
typedef long long LL;
const double eps = 1e-6;
const int INF = 2e9;
const LL LNF = 9e18;
const int MAXN = 1e6+10;

char s[MAXN];
int Next[MAXN];

void get_next(char s[], int m)
{
    int i, j;
    j = Next[0] = -1;
    i = 0;
    while(i<m)
    {
        while(j!=-1 && s[i]!=s[j]) j = Next[j];
        Next[++i] = ++j;
    }
}

int getmin(char *s, int len, int type) //返回最小表示法的始端
{
    int i = 0, j = 1, k = 0;
    while(i<len && j<len && k<len)
    {
        int t = s[(i+k)%len]-s[(j+k)%len];
        if (!t) k++;
        else
        {
            if (type*t>0) i += k+1;
            else j += k+1;
            if (i==j) j++;
            k = 0;
        }
    }
    return i<j?i:j;
}

int main()
{
    while(scanf("%s", s)!=EOF)
    {
        int len = strlen(s);
        int minex = getmin(s, len, 1);
        int maxex = getmin(s, len, -1);
        get_next(s, len);
        int r = len-Next[len];
        int times = len%r?1:len/r;

        printf("%d %d %d %d\n", minex+1, times, maxex+1, times);
    }
}