Number Sequence（KMP之子串首次出现的位置）

悠悠 2022-06-09 11:17 124阅读 0赞

# Number Sequence #

**Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 30033    Accepted Submission(s): 12666**

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

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
      
       
      
       
        
       题意：找出第二个数组在第一个数组中第一次出现的位置,位置从1开始；
      
       
      
       
       
        #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    int a[1000001],n,m;
    int b[1000001];
    int Next[1000005];
    void getnext()
    {
        int j=0,k=-1;
        Next[0]=-1;
        while(j<m-1)
        {
            if(k==-1||b[j]==b[k])
            {
                j++;
                k++;
                Next[j]=k;
            }
            else
                k=Next[k];
        }
    }
    int KMP()
    {
        int i=0,j=0;
        while(i<n&&j<m)
        {
            if(j==-1||a[i]==b[j])
            {
                i++;
                j++;
            }
            else
                j=Next[j];
        }
        if(j==m)
            return i-j+1;
        return -1;
    }
    int main()
    {
        int t;
        scanf("%d",&t);
        while(t--)
        {
            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]);
            getnext();
            int ans=KMP();
            printf("%d\n",ans);
        }
        return 0;
    }