HDU 3374 String Problem(最大最小表示法 模板题)

题目链接：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

abcderaaaaaaababab

 

Sample Output

1 1 6 11 6 1 61 3 2 3

 

Author

WhereIsHeroFrom

 

Source

HDOJ Monthly Contest – 2010.04.04

题意：

输出所给字符串的最大最小表示的起始位置和数量！

PS:

数量我们可以利用KMP里的求next数组来求的循环节！然后再套用一下最大最小表示法的模板就OK了！

代码如下：

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;#define MAXN 1000017int Next[MAXN];char str[MAXN];int k;void getNext( char T[],int len){    int i = 0, j = -1;    Next[0] = -1;    while(i < len)    {        if(j == -1 || T[i] == T[j])        {            i++,j++;            Next[i] = j;        }        else            j = Next[j];    }}//最小表示法int get_minstring(char *s){    int len = strlen(s);    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==0)            k++;        else        {            if(t > 0)                i+=k+1;            else                j+=k+1;            if(i==j) j++;            k=0;        }    }    return min(i,j);}//最大表示法int get_maxstring(char *s){    int len = strlen(s);    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==0)            k++;        else        {            if(t > 0)                j+=k+1;            else                i+=k+1;            if(i==j) j++;            k=0;        }    }    return min(i,j);}int main(){    while(scanf("%s",str)!=EOF)    {        int len = strlen(str);        getNext(str,len);        int tt = len - Next[len];        int num = 1;        if(len%tt == 0)        {            num = len/tt;        }        int posmin = get_minstring(str);        int posmax = get_maxstring(str);        printf("%d %d %d %d\n",posmin+1,num,posmax+1,num);    }    return 0;}