hdu1711 Number Sequence 求模式串在主串中的位置

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

213 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 1 313 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 2 1

Sample Output

6-1

#include<iostream>#include<cstring>#include<cstdio>#include<string>using namespace std;int s[1000010],t[10010];//s为主串，t为模式串int nextx[10010];//从1开始int slen,tlen;void get_next()////得到模式串str1的next[];{    int i=0,j=-1;    nextx[0]=-1;    while(i<tlen)    {        if(j==-1||t[i]==t[j])            nextx[++i]=++j;        else            j=nextx[j];    }}int kmp_index()////返回首次出现的位置,从0开始{    int j=0;    for(int i=0;i<slen;i++)    {       while(j!=-1&&s[i]!=t[j])            j=nextx[j];       j++;       if(j==tlen)           return i-tlen+2;    }    return -1;}int main(){    int T;    cin>>T;    while(T--)    {        cin>>slen>>tlen;        for(int i=0;i<slen;i++)            scanf("%d",&s[i]);        for(int i=0;i<tlen;i++)        scanf("%d",&t[i]);        get_next();        cout<<kmp_index()<<endl;    }    return 0;}