KMP

最新推荐文章于 2023-09-05 22:41:05 发布

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最新推荐文章于 2023-09-05 22:41:05 发布

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分类专栏：算法

本文链接：https://blog.youkuaiyun.com/wuxiao18/article/details/84569654

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本文深入探讨了KMP算法的工作原理，包括如何构建部分匹配表以及如何利用它避免不必要的字符比较，从而提高字符串搜索效率。通过实例解析，读者将能够理解和实现KMP算法。

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http://acm.hdu.edu.cn/showproblem.php?pid=1711

Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 42930 Accepted Submission(s): 17720

Problem Description

Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]. If there are more than one K exist, output the smallest one.

Input

The first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate a[1], a[2], ...... , a[N]. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000].

Output

For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.

Sample Input

13 5

1 2 1 2 3 1 2 3 1 3 2 1 2

1 2 3 1 3

13 5

1 2 1 2 3 1 2 3 1 3 2 1 2

1 2 3 2 1

Sample Output

6 -1

标程

#include<stdio.h>
#include<string.h>
int a[1000010],b[1000010],alen,blen;
int next[10010];
void get_next()
{
   int i,j;
   i=1;
   j=0;
   next[0]=0;
   while(i < blen)
   {
       if(j==0&&b[i]!=b[j])
       {
           next[i]=0;
           i++;
       }
       else if(j>0&&b[i]!=b[j])
           j=next[j-1];
       else
       {
           next[i]=j+1;
           i++;
           j++;
       }
   }
};

int main()
{
   int N,i,j,count;
   scanf("%d",&N);
   while(N--)
   {
       scanf("%d%d",&alen,&blen);
       for(i = 0;i < alen;i ++)
           scanf("%d",&a[i]);
       for(i = 0;i < blen;i ++)
           scanf("%d",&b[i]);
           get_next();
       i = 0;
       j = 0;
       count = -1;
       while(i < alen)
       {
           if(j == 0 && a[i] != b[j])
               i++;
           else if(j > 0 && a[i] != b[j])
               j = next[j-1];
           else
           {
               i ++;
               j ++;
           }
           if(j == blen)
           {
               count = i-blen+1;
               break;
           }
       }
       printf("%d\n",count);
   }
   return 0;
}