Cyclic Nacklace HDU - 3746

制作魅力手链

最新推荐文章于 2019-07-27 20:27:36 发布

原创最新推荐文章于 2019-07-27 20:27:36 发布 · 381 阅读

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本文介绍了一种通过添加最少数量的彩色珍珠来将普通手链转化为循环结构的魅力手链的方法。文章详细解释了如何找到最小循环节，并给出了具体的算法实现。

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

讲一下我对循环节的理解：

注：下面所有的大写字母ABCD...代表区间，不是真实的字符。

我们考虑整个串的前缀和后缀的最大匹配，也就是 f[len], 也就是上边的第一个字符串的后缀BA和第二个字符串的CB。

首先，我们可以假设，len-f[len] 为d,也就是 A的长度，那么我们可以按照 d,这个长度把整个字符串分成若干份，如果有冗余我们不考虑那一段。

然后，我们按照d的长度对整个字符串从后往前，进行标号，可以发现A=B，B=C（前缀和后缀的相对位置相同），....,可以一直到最后一个，所以这种划分是可行的。

最后，怎么知道这个d是最小的呢，如果d不是最小的，说明，后缀和前缀的匹配长度还可增加，也就是上边的串可以往右移，可是f[len]已经是最大长度，上边的串不可以再移动，所以d就是最小的循环节。

所以循环节的原理差不多就是这样了，至于添加几个，可以直接考虑最前面的冗余，然后 loop-余数，就是答案了。至于题目说的可以从后面或者前面添加，也就是，我们也可以反过来考虑，从前面开始划分。

#include<stdio.h>
#include<string.h>
const int maxn=1e5+10;
char str[maxn];
int fail[maxn],f[maxn];

void getFail(char* P,int len){

    fail[0]=fail[1]=0;
    f[0]=f[1]=0;
    for(int i=1;i<len;i++){
        int j=f[i];
        while(j&&P[i]!=P[j])j=f[j];
        fail[i+1]=f[i+1]=(P[i]==P[j]?j+1:0);
        if(fail[i+1]==j+1&&P[i+1]==P[j+1])fail[i+1]=fail[j+1];
    }
}
//推荐刘汝佳 白书Kmp的状态机理解

int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%s",str);
        int len=strlen(str);
        getFail(str,len);

        //for(int i=0;i<=len;i++)printf("%d%c",fail[i],i==len?'\n':' ');
        //for(int i=0;i<=len;i++)printf("%d%c",f[i],i==len?'\n':' ');

        int loop=len-fail[len];
        if(len>loop&&len%loop==0){
            printf("0\n");
        }
        else{
            printf("%d\n",loop-len%loop);
        }
    }
    return 0;
}