hdu1711 Number Sequence 【kmp】

最新推荐文章于 2019-07-14 13:24:47 发布

Vace___yun

最新推荐文章于 2019-07-14 13:24:47 发布

阅读量236

点赞数

CC 4.0 BY-SA版权

分类专栏： kmp 文章标签： kmp

本文链接：https://blog.youkuaiyun.com/Vace___yun/article/details/89203362

kmp 专栏收录该内容

0 篇文章

订阅专栏

本文深入解析了KMP算法，一种高效的字符串匹配算法，用于在主串中查找模式串出现的位置。通过实例说明了如何实现KMP算法，包括计算next数组和进行匹配的过程，适用于解决大规模文本搜索问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

2
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

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int n,N,M;
int a[1000010],b[1000010],p[1000010];
int len1,len2;
void getnext()
{
    int i=0,j=-1;
    p[0]=-1;
    while(i<M)
    {
        if(j==-1||b[i]==b[j])
        {
            i++;
            j++;
            p[i]=j;
        }
        else
            j=p[j];
    }
}
int kmp()
{
    int i=0,j=0;
    getnext();
    while(i<N)
    {
        if(j==-1||a[i]==b[j])
        {
            i++;
            j++;
        }
        else
            j=p[j];
        if(j==M)
            return i-j+1;
    }
    return -1;
}
int main()
{
    scanf("%d",&n);
    while(n--)
    {
        memset(p,0,sizeof(p));
        scanf("%d%d",&N,&M);
        for(int i=0;i<N;i++)
            scanf("%d",&a[i]);
        for(int i=0;i<M;i++)
            scanf("%d",&b[i]);
        printf("%d\n",kmp());
    }
    return 0;
}