String Problem HDU - 3374 （KMP+最小表示法）

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本文探讨了字符串经过左移操作生成的所有可能字符串中，如何找到字典序最小和最大的字符串，并计算其出现次数。通过使用KMP算法确定最小循环节，解决了这一问题。

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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

题目大意：对于每个字符串输出以哪个开始的字符串字典序是最小的，再输出有几个循环节，再输出字典序最大的，在输出有几个循环节。

解题思路：用KMP找出最小循环节，对于字典序用最小表示法即可。

/*
@Author: Top_Spirit
@Language: C++
*/
#include <bits/stdc++.h>
using namespace std ;
typedef unsigned long long ull ;
typedef long long ll ;
const int Maxn = 1e6 + 10;
const int INF = 0x3f3f3f3f ;
const ull seed = 133 ;
const int MOD = 10007 ;
const double PI = acos(-1.0) ;
const int Max = 200 + 10 ;

string s ;
int len ;
int Next[Maxn] ;

void getNext(){
    Next[0] = -1 ;
    int j = 0, k = -1 ;
    while (j < len){
        if (k == -1 || s[j] == s[k]) Next[++j] = ++k ;
        else  k = Next[k] ;
    }
}

int Min_or_Max(int flag){
    int i = 0, j = 1, k = 0 ;
    while (i < len && j < len && k < len){
        int tmp = s[(i + k) % len] - s[(j + k) % len] ;
        if (!tmp) k++ ;
        else if (!flag){
            if (tmp < 0) i += k + 1;
            else j += k + 1 ;
            if (i == j) j++ ;
            k = 0 ;
        }
        else if (flag){
            if (tmp > 0) i += k + 1 ;
            else j += k + 1 ;
            if (i == j) j++ ;
            k = 0;
        }
    }
    return min(i, j) ;
}

int main (){
    ios_base::sync_with_stdio(false) ;
    cin.tie(0) ;
    cout.tie(0) ;
    while (cin >> s){
        memset(Next, 0, sizeof(Next)) ;
        len = s.size() ;
        getNext() ;
        int Max = Min_or_Max(0) ;
        int Min = Min_or_Max(1) ;
        int cnt = len - Next[len] ;
        int ans = len % cnt == 0 ? len / cnt : 1;
        cout << Min + 1 << " " << ans << " " << Max + 1 << " " << ans << endl ;
    }
    return 0;
}