hdoj 1711 Number Squence

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

 

对于kmp，最重要的是理解next数组（  反映出现失配情况时，应该跳过多少无用字符（也即模式串应该向右滑动多长距离）而进行下一次检测）

模式串的研究是重点，和匹配串无关系，next的下标表示模式串与匹配串匹配到位数，其下标对应的值是该位前模式串max（前缀=后缀）的个数，

 

 这个算出的next数组要减一

刚接触，懵懂（b站视频里面有讲，看了三个人的视频，依旧懵懂，(╥╯^╰╥)(╥╯^╰╥)慢慢来，加油！）

http://billhoo.blog.51cto.com/2337751/411486      这位大佬讲的也不错

#include<iostream>
    #include <string.h>
    using namespace std;
    int t[10005],s[1000005],Next[10005];
    void getnext(int t[],int Next[],int n)
    {
        int i=1;
        int j=0;
        Next[1]=0;
        while(i<n)
        {
            if(j==0||t[i-1]==t[j-1])
            {
                i++;
                j++;
                Next[i]=j;
            }
            else
                j=Next[j];
        }
    }
    int kmp(int t[],int s[],int Next[],int a,int b)
    {
        int i=1;
        int j=1;
        while(i<=a&&j<=b)
        {
            if(j==0||s[i-1]==t[j-1])
            {
                i++;
                j++;
            }
            else
                j=Next[j];
        }
        if(j>b)
            return i-b;
        else
            return 0;
    }
    int main()
    {
        int n;
        int a,b;
        cin>>n;
        while(n--)
        {
            cin>>a>>b;
            for(int i=0;i<a;i++)
                cin>>s[i];
            for(int j=0;j<b;j++)
                cin>>t[j];
                memset(Next,0,sizeof(Next));
            getnext(t,Next,b);
            int result=kmp(s,t,Next,a,b);
            if(result==0)
                cout<<"-1"<<endl;
            else
                cout<<result<<endl;
        }
        return 0;
    }