HDU 3374 String Problem（最小表达法+kmp）

String Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3328    Accepted Submission(s): 1372

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

abcderaaaaaaababab

 

Sample Output

1 1 6 11 6 1 61 3 2 3

 

Author

WhereIsHeroFrom

 

Source

HDOJ Monthly Contest – 2010.04.04 

 

题意：

给你一个字符串，让你求出字典序最小的同构字符串，并且求出有几个同样的。

同理在求最大的。

POINT：

最小表示法

http://blog.csdn.net/zy691357966/article/details/39854359

#include <iostream>#include <string.h>#include <stdio.h>using namespace std;const int N =1000000+88;int nxt[N];void prenxt(char s[]){    int l=strlen(s);    int i=0,j;    j=nxt[0]=-1;    while(i<l)    {        while(j!=-1&&s[j]!=s[i]) j=nxt[j];        nxt[++i]=++j;    }}int findmin(char s[]){    int i=0,j=1,k=0;    int len=strlen(s);    while(i<len&&j<len&&k<len)    {        int t=s[(i+k)%len]-s[(j+k)%len];        if(t==0) k++;        else        {            if(t>0) i+=k+1;            else j+=k+1;            if(i==j) j++;            k=0;        }    }    return min(i,j);}int findmax(char s[]){    int i=0,j=1,k=0;    int len=strlen(s);    while(i<len&&j<len&&k<len)    {        int t=s[(i+k)%len]-s[(j+k)%len];        if(t==0) k++;        else        {            if(t>0) j+=k+1;            else i+=k+1;            if(i==j) j++;            k=0;        }    }    return min(i,j);}char s[N];int ans1,ans2;int doit(int pos){    int l=strlen(s);    prenxt(s);    int ans=1;    if(l%(l-nxt[l])==0&&nxt[l]!=0)        ans=l/(l-nxt[l]);    return ans;}int main(){    while(~scanf("%s",s))    {        memset(nxt,0,sizeof nxt);        int pos1=findmin(s);        ans1=doit(pos1);        int pos2=findmax(s);        ans2=doit(pos2);        printf("%d %d %d %d\n",pos1+1,ans1,pos2+1,ans2);                    }}