HDU-3746-Cyclic Nacklace(KMP，循环节）

Posted 2021-01-04 1625--h

Cyclic Nacklace

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 15422 Accepted Submission(s): 6425

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

分析

• 题意：将所给的字符串变成多个完整的循环（至少两个），然后给出最少需要添加的字符数。
• 必须找出最小循环节，才可以求出满足题意中的最小添加数。
• KMP算法中，可以利用k = len - nxt[len]来求出最小循环节长度k。
  • KMP算法中匹配串b指针j的移动可以理解为长度为最小循环节的移动。
• 然后只需要判断在最小循环节数大于1的情况下，串是否是由整数个最小循环子串构成。如果是，就不需要添加。
• 如果需要添加，则只需要串长对最小循环节数取余求出多余长度，然后最小循环长度减去这个多余长度，即可求出需要添加数目。

    string a;
    int n;
    int nxt[100001];
    void getnxt()
    {
        nxt[0] = -1;
        int j = 0,k = -1;
        while(j<n)
        {
            if(k==-1||a[j] == a[k])
                nxt[++j] = ++k;
            else
                k = nxt[k];
        }
    }
    int main() 
    {
        int t;cin>>t;
        while(t--)
        {
            cin>>a;
            n = a.length();
            getnxt();
            int k = n-nxt[n];//k为最小循环节长度
            if(n%k==0&&k!=n)cout<<0<<endl;//不需要添加
            else
                cout<<k-nxt[n]%k<<endl;
        }
        return 0;
    }