poj1509 字符串最小表示法

Glass Beads

Time Limit: 3000MS Memory Limit: 10000KTotal Submissions: 2708 Accepted: 1573

Description

Once upon a time there was a famous actress. As you may expect, she played mostly Antique Comedies most of all. All the people loved her. But she was not interested in the crowds. Her big hobby were beads of any kind. Many bead makers were working for her and they manufactured new necklaces and bracelets every day. One day she called her main Inspector of Bead Makers (IBM) and told him she wanted a very long and special necklace. 

The necklace should be made of glass beads of different sizes connected to each other but without any thread running through the beads, so that means the beads can be disconnected at any point. The actress chose the succession of beads she wants to have and the IBM promised to make the necklace. But then he realized a problem. The joint between two neighbouring beads is not very robust so it is possible that the necklace will get torn by its own weight. The situation becomes even worse when the necklace is disjoined. Moreover, the point of disconnection is very important. If there are small beads at the beginning, the possibility of tearing is much higher than if there were large beads. IBM wants to test the robustness of a necklace so he needs a program that will be able to determine the worst possible point of disjoining the beads. 

The description of the necklace is a string A = a1a2 ... am specifying sizes of the particular beads, where the last character am is considered to precede character a1 in circular fashion. 

The disjoint point i is said to be worse than the disjoint point j if and only if the string aiai+1 ... ana1 ... ai-1 is lexicografically smaller than the string ajaj+1 ... ana1 ... aj-1. String a1a2 ... an is lexicografically smaller than the string b1b2 ... bn if and only if there exists an integer i, i <= n, so that aj=bj, for each j, 1 <= j < i and ai < bi

Input

The input consists of N cases. The first line of the input contains only positive integer N. Then follow the cases. Each case consists of exactly one line containing necklace description. Maximal length of each description is 10000 characters. Each bead is represented by a lower-case character of the english alphabet (a--z), where a < b ... z.

Output

For each case, print exactly one line containing only one integer -- number of the bead which is the first at the worst possible disjoining, i.e.\ such i, that the string A[i] is lexicographically smallest among all the n possible disjoinings of a necklace. If there are more than one solution, print the one with the lowest i.

Sample Input

4helloworldamandamandadontcallmebfuaaabaaa

Sample Output

101165

题意：给你一串字符串  构成一个环 问你从哪个位置剪短后  使这个字符串在字典序中最小

思路：字符串的最小表示法

算法解释：

   
(1).利用两个指针i，j。初始化时i指向s[0]，j指向s[1]。(2).k=0开始，检验s[i+k]和s[j+k]是否相等，相等则k++，一直下去，直到找到第一个不相同的字符（若k试了一个字符串的长度也没找到不同，即整个串都是相同的字符。则那个位置就是最小表示位置，算法终止并返回）。该过程中s[i+k]和s[j+k]的关系有三种：　　1).s[i+k]>s[j+k]，i滑动到i+k+1处，s[i-i+k-1]不会是循环字符串的"最小表示"的前缀。
　　2).s[i+k]<s[j+k]，j滑动到j+k+1处。　　3).s[i+k]==s[j+k]，则k++，if(k==len)返回结果。　　若滑动后i==j，将正在变化的那个指针在+1.直到i，j把整个字符串都检验完毕，返回两者中小于len的值。(3).返回min( i , j )
模板：
 int Min(char *s,int l)//串s[0~l-1]的最小表示位置
{
    int i = 0, j = 1, k = 0,t;
    while (i < l && j < l && k < l)
    {
        t = s[(i+k)%l] - s[(j+k)%l];
        if (t == 0)
            k++;//相等的话,检测长度加1
        else 
        {
            if (t > 0)
                i += k + 1;
            else
                j += k + 1;
            if (i == j)
                j++;//或者i++
            k = 0;
        } 
    } 
    return min(i,j);
}
顺便附上最大表示法代码
int Max(int len,char pat[])  //最大表示法
{
   int i=0,j=1,k=0;
   while(i<len && j<len && k<len)
   {
       int t = pat[(i+k)%len] - pat[(j+k)%len];
       if(!t) k++;
       else
       {
           if(t>0) j = j+k+1;
           else i = i+k+1;
           if(i == j) j++;
           k = 0 ;
       }
   }
   return i<j?i:j;
}