KMP

由本人喜欢用题来进行讲解所以就在这里用题进行相关的讲解

这里放一两个大神的讲解1:https://blog.youkuaiyun.com/v_july_v/article/details/7041827

                                       2:https://blog.youkuaiyun.com/Dch19990825/article/details/82764195

相信看完之后你应该知道next到底是什么了

Number Sequence

 

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：找到第第一行的字符串中与第二行的字符串相照应的位置，输出它，如果出现的不止一次输出最开始出现的位置，如果没有，输出-1

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int n,m,a[1000005],b[10005],nnext[10005];
/*next[i]的含义是字符串第i项之前的公共前后缀的长度*/
void getnext()
{
    int i=0,j=-1;
    nnext[0]=-1;
    while(i<m)
    {
        if(j==-1||b[i]==b[j])
        {
            i++,j++;
                nnext[i]=j;
        }
        else
            j=nnext[j];
    }
}
int kmp()
{
    getnext();
    int i=0,j=0;
    while(i<n&&j<m)
    {
        if(j==-1||a[i]==b[j])
        {
            i++,j++;
            if(j==m)
                return i-j+1;
        }
        else
            j=nnext[j];
    }
        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]);
        printf("%d\n",kmp());
    }
}