【1711 KMP】

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8883    Accepted Submission(s): 4086

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

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

 

/*********************** author:maple_小贾* Pro:HDOJ 1711* algorithm: KMP  使用全局变量* date:2013/11/8***********************/#include<cstdio>#include<cstring>#define MAXN 1000000+5int next[MAXN],s[MAXN],p[MAXN];int n,m;//主串和模式串的串长void GetNext(int *p, int *next){    int i,j;    next[0]=-1;    i=0,j=-1;    while(i<m){        if(j==-1 || p[i]==p[j]){//如果p[j]==p[i];            i++;            j++;            next[i]=j;        }        else {//不匹配 p][j]!=p[i];            j=next[j];        }    }}int KMP(int *s,int *p){    GetNext(p,next);    int i,j;    i=0,j=0;    while(i<n){        if(j==-1 || s[i]==p[j]){            i++;            j++;        }        else{            j=next[j];        }        if(j==m)            return i-m+1;    }    return -1;}int main(){    int test;    scanf("%d",&test);    while(test--){        scanf("%d%d",&n,&m);        for(int i=0;i<n;i++)            scanf("%d",&s[i]);        for(int i=0;i<m;i++)            scanf("%d",&p[i]);        printf("%d\n",KMP(s,p));    }}