HDU 3746 Cyclic Nacklace(kmp next数组运用)

Cyclic Nacklace

Problem Description

CC always becomes very depressed at the end of this month, he has checked his credit card yesterday, without any surprise, there are only 99.9 yuan left. he is too distressed and thinking about how to tide over the last days. Being inspired by the entrepreneurial spirit of "HDU CakeMan", he wants to sell some little things to make money. Of course, this is not an easy task.

As Christmas is around the corner, Boys are busy in choosing christmas presents to send to their girlfriends. It is believed that chain bracelet is a good choice. However, Things are not always so simple, as is known to everyone, girl's fond of the colorful decoration to make bracelet appears vivid and lively, meanwhile they want to display their mature side as college students. after CC understands the girls demands, he intends to sell the chain bracelet called CharmBracelet. The CharmBracelet is made up with colorful pearls to show girls' lively, and the most important thing is that it must be connected by a cyclic chain which means the color of pearls are cyclic connected from the left to right. And the cyclic count must be more than one. If you connect the leftmost pearl and the rightmost pearl of such chain, you can make a CharmBracelet. Just like the pictrue below, this CharmBracelet's cycle is 9 and its cyclic count is 2:

Now CC has brought in some ordinary bracelet chains, he wants to buy minimum number of pearls to make CharmBracelets so that he can save more money. but when remaking the bracelet, he can only add color pearls to the left end and right end of the chain, that is to say, adding to the middle is forbidden.
CC is satisfied with his ideas and ask you for help.

 

Input

The first line of the input is a single integer T ( 0 < T <= 100 ) which means the number of test cases.
Each test case contains only one line describe the original ordinary chain to be remade. Each character in the string stands for one pearl and there are 26 kinds of pearls being described by 'a' ~'z' characters. The length of the string Len: ( 3 <= Len <= 100000 ).

 

Output

For each case, you are required to output the minimum count of pearls added to make a CharmBracelet.

 

Sample Input

3

aaa

abca

abcde

 

Sample Output

0

2

5

题目大意：给定一个字符串，问你最少需要添加多少个‘a'~'z'的字符使它成为一个循环的字符串

思路：kmp，由kmp我们可以知道next数组的作用是求得子串s每个位置的相同前缀后缀的个数，不过我这里的nex数组是从{-1，0，1，2} 分别代表相同个数有{0，1，2，3}这么多个，那么我们可以看到next[len]也就是整个子串的相同前后缀个数，如果该个数等于0，则说明要想构成一个循环还需要一个子串的长度，所以此时需要len个，其次如果该个数==len-1 就说明子串已经构成了循环不需要在添加字符，最后只有部分相同的话len-nex[len]就是他的最小循环节，需要加上len-nex[len] - len%(len-nex[len])..不过后面看到一篇博客给出了kmp字符串循环节的解释，感觉emm超简洁 哈哈哈哈 大家可以去学学 http://www.cnblogs.com/wuyiqi/archive/2012/01/06/2314078.html

 

#include<iostream>
#include<algorithm>
#include<string>

using namespace std;
const int maxn = 100005;
int nex[maxn], T;
void Getnext(string a, int len)
{
    nex[0] = -1; int k = -1;
    for (int i = 1; i < len; i++) {
        while (k > -1 && a[k + 1] != a[i])k = nex[k];
        if (a[k + 1] == a[i])k++;
        nex[i] = k;
    }
}
int main()
{
    ios::sync_with_stdio(false);
    cin >> T;
    string str;
    while (T--) {
        cin >> str; int len = str.length();
        Getnext(str, len);
        int k = len - (nex[len - 1] + 1);
        if (len%k == 0 && len != k)cout << "0" << endl;
        else cout << k - len % k << endl;
    }
    return 0;
}