KMP(看毛片)-

KMP  (看毛片)哈哈 ,  字符匹配问题,   说实话 网上的版本很多, 很多大牛都写了自己的理解, 看了又看, 迷迷糊糊的 ,

确实不好理解 特别是next[] 数组

附上 几个大牛们的 KMP讲解 

  KMP 模板(刘汝佳算法入门):  http://blog.csdn.net/liuzhushiqiang/article/details/8941032

  KMP 讲解   (这个不错) :  http://blog.csdn.net/jjdiaries/article/details/12689413

http://blog.csdn.net/joylnwang/article/details/6778316

我们可以模拟一下 这个步骤,     next[ ]  每次存的是 当前字符的前一个字符在与前面字符 对比中  要跳转的 个数

 KMP  会有一个预处理  计算next 跳转数 

 预处理操作

    int len = strlen(s);    int i,j=0; next[0] = next[1] = 0;for(i = 2; i <=len; i++)    {while(j>0&&s[j]!=s[i-1]) j=next[j];if(s[j]==s[i-1]) j++; next[i]=j;}

 然后就是 字符匹配  一个模板  另一个则是 要匹配的 串

     void kmp_to_kmp(string a,string b,int *next)     {        //a 母板 b 带匹配的串         kmp_self(b,next);// 预处理B 串 更新next 跳转数        int i,j=0;        for(i=0;i<a.length();i++)     {         while(j>0&&b[j]!=a[i])              j=next[j-1];         if(b[j]==a[i])             j++;         if(j==b.length())         {             printf("YES  %d \n ",(i-b.length()+1));             return ;         }     }     printf("NO\n");}

Cyclic Nacklace

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8614    Accepted Submission(s): 3703

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

3aaaabcaabcde

 

Sample Output

025

 

Author

possessor WC

 

题目意思  :   问你 输入串后  需要  再 加多少个 字母 后  让它变成 重复的 例如 aaa 则不需要  abca---> abcabc    abcde-->> abcdeabcde

 

 只需要预处理操作, 计算最小循环节 就可以  

#include <iostream>#include <string>#include <algorithm>#include <cstring>#include <stdio.h>using namespace std;void kmp_self(char *s,int *next)// 自己匹配自己{    int len = strlen(s);    int i,j=0; next[0] = next[1] = 0;for(i = 2; i <=len; i++)    {while(j>0&&s[j]!=s[i-1]) j=next[j];if(s[j]==s[i-1]) j++; next[i]=j;}    int cir=len-next[len];//最小循环节    if(cir!=len&&len%cir==0)        printf("0\n");    else        printf("%d\n",cir-next[len]%cir);}int main(){    int T;    int i,j;    char s[100010];    int next[100010];    scanf("%d",&T);    {        getchar();        while(T--)        {            scanf("%s",s);// 一定要注意   没有memset(s) 操作 否则必 W A              memset(next,0,sizeof(next));            kmp_self(s,next);        }    }    return 0;}