KMP 匹配个数

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这道题是我做的第一道KMP的题目，还是依赖于叶大佬帮我改的，在此我之前的文章中曾经记录过KMP求解匹配字符串的位置，这次是匹配出有几个字符串，其实这道题不用KMMP也可以做的，，，

下面是题目：

One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems.

But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name.

It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita".

Names are case sensitive.

Input

The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 — the name of the problem.

Output

Print "YES", if problem is from this contest, and "NO" otherwise.

Examples
input
Alex_and_broken_contest
output
NO
input
NikitaAndString
output
YES
input
Danil_and_Olya
output
NO

题意就是母串中仅能包括一个字串是YES，否则是NO。

接下来是我用KMP做出来的代码：

#include <stdio.h>#include <string.h>int next[1002] = {-5};char v[10][10] = {"Nikita","Danil","Olya","Slava","Ann"};char T[1000]; void get_next(int len2){int k = -1;int j = 0;next[j] = k;while(j<len2){if((k==-1)||(v[j] == v[k]))//判断匹配成功的条件 {k++;j++;next[j] = k;//匹配成功都往后移 }else{k = next[k];//匹配不成功，看j串是第几个不成功，假设第n个不成功，即j回溯为next[n-1]，找n-1处//的next值，即为k的下一次的值，然后下一次从k的值处开始匹配。 }}}int index_KMP(int len2,int len1,int pos,int l){int i,j;int ans = 0;i = pos, j = 0;while(i<len1){if((j==-1)||(T[i] == v[l][j])){++i;++j;}else{j = next[j];}if(j == len2){ans++;j = next[j-1];i--; }}//printf("%d\n",ans);return ans;}int main(){scanf("%s",T);int len1 = strlen(T);int pos = 0;int index = 0;for(int i=0;i<5;i++){int len2 = strlen(v[i]);get_next(len2);}for(int i=0;i<5;i++){int len2 = strlen(v[i]);index += index_KMP(len2,len1,0,i);}//printf("%d\n",index);if(index==1)printf("YES\n");elseprintf("NO\n");}

接下来是我从一篇大佬的博客中学会的KMP如何匹配出子串的个数的模板：

#include<cstdio>  #include<cstring>  using namespace std;    const int MAXW=10001,MAXT=1000001;  char W[MAXW],T[MAXT];  int next[MAXW];  int lenW,lenT;    void getnext(int lenW)  {      int i=0,j=-1;      next[i]=-1;      while(i<lenW) {          if(j==-1||W[i]==W[j]) {              next[++i]=++j;          }          else j=next[j];      }  }    int kmp(int pos,int lenW,int lenT)  {      int i=pos,j=0,ans=0;      while(i<lenT) {          if(T[i]==W[j]||j==-1) ++i,++j;          else j=next[j];          if(j==lenW) {              ans++;              j=next[j-1];              i--;          }      }      return ans;  }      int main()  {      int n;      scanf("%d",&n);      while(n--) {          scanf("%s%s",W,T);          lenW=strlen(W);          lenT=strlen(T);          getnext(lenW);          printf("%d\n",kmp(0,lenW,lenT));      }      return 0;  }

继续努力吧！

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