hdu——1711——Number Sequence

最新推荐文章于 2019-10-31 13:52:12 发布

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最新推荐文章于 2019-10-31 13:52:12 发布

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分类专栏： kmp HDU

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本文介绍了一种高效的字符串匹配算法——KMP算法，并通过一个具体的编程实例详细展示了如何使用KMP算法解决寻找子串的问题。该算法首先通过构建部分匹配表(next数组)来避免重复比较，从而达到线性时间复杂度。

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

#include <iostream>
using namespace std;
int next[100000];
int n,m;
int s1[1000000];
int s2[10000];
void getnext()
{
    int j=0;
    int k=-1;
    next[0]=-1;
    while(j<m-1)
    {
        if(k==-1||s2[j]==s2[k])
        {
           j++;
           k++;
           next[j]=k;
        }
        else
            k=next[k];
    }
}
int kmp()
{
    int i=0;
    int j=0;

    while(i<n)
    {
        if(j==-1||s1[i]==s2[j])
        {
            i++;
            j++;
        }
        else
        {
            j=next[j];
        }
        if(j==m)
        {
            return i-m+1;
        }
    }
    return -1;
}
int main()
{
    int k;
    cin>>k;
    while(k--)
    {
        cin>>n>>m;
        for(int i=0;i<n;i++)
        {
            //cin>>s1[i];
            scanf("%d",&s1[i]);
        }
        for(int i=0;i<m;i++)
        {
            //cin>>s2[i];
            scanf("%d",&s2[i]);
        }
        getnext();
        if(n<m)
            cout<<"-1"<<endl;
        else
            cout<<kmp()<<endl;
    }
    return 0;
}