【KMP】 hdu3374 String Problem

String Problem

http://acm.hdu.edu.cn/showproblem.php?pid=3374

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
   
   
   
   

最小最大表示法+ KMP

求最小表示和最大表示的开始位置（多个解输出最前面的那个）与表示的次数

最小表示法 http://blog.csdn.net/cclsoft/article/details/5467743

#include<cstdio>
#include<cstring>
using namespace std;
#define min(a,b) ((a)<(b)?(a):(b))
char s[1000005];
int next[1000005];
void get_next()
{
    int i=0,j=-1;
    next[0]=-1;
    for(; s[i];)
        if(j==-1||s[i]==s[j])
        {
            ++i;
            ++j;
            next[i]=j;
        }
        else  j=next[j];
}
int MaxmumRepresentation(int l)//最大表示法
{
    int i=0,j=1,k=0,t;
    for(;i<l&&j<l&&k<l;)
    {
        t=s[(i+k)%l]-s[(j+k)%l];
        if(!t) ++k;
        else
        {
            if(t<0) i=i+k+1;
            else j=j+k+1;
            if(i==j) ++j;
            k=0;
        }
    }
    return min(i,j);
}
int MinmumRepresentation(int l)//最小表示法
{
    int i=0,j=1,k=0,t;
    for(;i<l&&j<l&&k<l;)
    {
        t=s[(i+k)%l]-s[(j+k)%l];
        if(!t) ++k;
        else
        {
            if(t>0) i=i+k+1;
            else j=j+k+1;
            if(i==j) ++j;
            k=0;
        }
    }
    return min(i,j);
}
int main()
{
    for(; ~scanf("%s",s);)
    {
        int len=strlen(s);
        int posmax=MaxmumRepresentation(len);
        int posmin=MinmumRepresentation(len);
        get_next();
        int r=len-next[len];//kmp求循环节长度
        if(len%r==0) r=len/r;
        else         r=1;
        printf("%d %d %d %d\n",posmin+1,r,posmax+1,r);
    }
    return 0;
}