【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

   
   
   
   

    
    
    
    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
   
   
   
   

 

/**********************
* 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+5
int 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));
    }
}