HDU-3374

String Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2057    Accepted Submission(s): 897

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

/**
    题意：给一个串，然后求串的最小表示法和最大表示法，并且输出有几个
    做法：串的最小表示法 + KMP
**/
#include <iostream>
#include <algorithm>
#include <cmath>
#include <string.h>
#include <stdio.h>
#include <queue>
#include <map>
#define maxn 1000005
using namespace std;
int mmin, mmax;
int len;
char t1[maxn];
char str[maxn << ];
char t2[maxn];
void kmp_pre(char x[], int m, int nxt[])
{
    int i, j;
    j = nxt[] = -;
    i = ;
    while(i < m)
    {
        while(- != j && x[i] != x[j]) {
            j = nxt[j];
        }
        nxt[++i] = ++ j;
    }
}
int nxt[maxn];
int kmp_count(char x[], int m, char y[], int n)
{
    int i, j;
    int ans =  ;
    kmp_pre(x, m, nxt);
    i = j = ;
    while(i < n)
    {
        while(- != j && y[i] != x[j]) {
            j = nxt[j];
        }
        i++;
        j++;
        while(j >= m)
        {
            ans ++;
            j = nxt[j];
        }
    }
    return ans;
}
void matchmin(char  s[])
{
    int i, j, k, l;
    int N = len;
    for(i = , j = ; j < N;)
    {
        for(k = ; k < N && s[i + k] == s[j + k]; k++);
        if(k >= N) {
            break;
        }
        if(s[i + k] < s[j + k]) {
            j += k + ;
        }
        else
        {
            l = i + k;
            i = j;
            j = max(l, j) + ;
        }
    }
    mmin = i + ;
}
void  matchmax(char s[])
{
    int i, j, k, l;
    int N = len;
    for(i = , j = ; j < N;)
    {
        for(k = ; k < N && s[i + k] == s[j + k]; k++);
        if(k >= N) {
            break;
        }
        if(s[i + k] > s[j + k]) {
            j += k + ;
        }
        else
        {
            l = i + k;
            i = j;
            j = max(l, j) + ;
        }
    }
    mmax = i + ;
}
int main()
{
    while(~scanf("%s", str))
    {
        int n;
        string tmp;
        len = strlen(str);
        for(int i = ; i < len; i++)
        {
            str[i + len] = str[i];
        }
        matchmin(str);
        matchmax(str);
        for(int i = ; i < len; i++)
        {
            t1[i] = str[i + mmin - ];
            t2[i] = str[i + mmax - ];
        }
        int res1 = kmp_count(t1, len, str, len + len);
        int res2 = kmp_count(t2, len, str, len + len);
        if(mmin == ) {
            res1 --;
        }
        if(mmax == ) {
            res2 --;
        }
        printf("%d %d %d %d\n", mmin, res1, mmax, res2);
    }
    return ;
}