Cyclic Nacklace HDU - 3746 (kmp)

Cyclic Nacklace

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15406    Accepted Submission(s): 6419

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

 

Author

possessor WC

 

 

Source

HDU 3rd “Vegetable-Birds Cup” Programming Open Contest 

 

 

Recommend

lcy

 

题意：给了你一个由英文字母组成的一个类似于项链的串，问最少添加几颗珠子可以使得它是一个循环，例如aaa不用添加就可以循环，abca只需要加两颗珠子bc就可以组成abcabc使得它循环，而abcde只能添加五颗珠子使得abcdeabcde为循环

思路：kmp题目，不需要计数和位置，其实就是next数组的一个应用，显然最小的环就是len-next[len]，所以我们得到周期t后就可以考虑添加珠子的数量了，如果周期和原来的珠子相同，说明它是类似于abcde的这种周期就是它本身的这种，如果本身的长度是周期的倍数，当然这里不可能是一倍，一倍就是本身就是周期的情况了， 所以像abcabc，aaa这种就不需要在加珠子了，最后一种就是还有需要加使得它构成一个周期的，比如aabbaabbaa，显然周期aabb，所以我们只需要在末尾加上bb就OK了，我们将bb接到aa后面就再次构成了一个周期。代码其实很短，比模版短多了，但对于next数组要理解，做这道题千还专门去hihocoder看了看next数组，然后草稿纸走了一遍程序，又输出了一下next数组求解的递归过程，呜呜呜，好菜啊

 

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<set>
#include<vector>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-10
#define PI acos(-1.0)
#define ll long long
const int maxn=1000005;

void kmp_pre(char x[],int m,int nex[])
{
    int i,j;
    j=nex[0]=-1;
    i=0;
    while(i<m)
    {
        while(-1!=j && x[i]!=x[j])
            j=nex[j];
        nex[++i]=++j;
    }
}

int nex[maxn];

int main()
{
    int t;
    char str[maxn];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",str);
        int len=strlen(str);
        kmp_pre(str, len, nex);
        int whi=len-nex[len];
        if(whi==len)
            printf("%d\n",len);
        else if(len % whi==0)
            printf("0\n");
        else
            printf("%d\n",whi-len%whi);
        
    }
}

 

转载于:https://www.cnblogs.com/smallhester/p/9582961.html