kmp

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
  1. #include<stdio.h>  
  2. #include<string.h>  
  3. int next[10005];  
  4. int s[1000005],p[10005];  
  5. int n,m;  
  6. void getNext(int p[])  
  7. {  
  8.     int i=0,j=-1,len=m-1;  
  9.     next[0]=-1;  
  10.     while(i<len)  
  11.     {  
  12.         if(j==-1||p[i]==p[j])  
  13.         {  
  14.             i++,j++;  
  15.             next[i]=j;}  
  16.         else j=next[j];  
  17.     }  
  18. }  
  19. int kmp(int n,int m)  
  20. {  
  21.     int i=0,j=0;  
  22.     while(i<n&&j<m)  
  23.     {  
  24.         if(j==-1||s[i]==p[j])  
  25.         i++,j++;  
  26.         else j=next[j];  
  27.     }  
  28.     if(j==m)return i-j+1;  
  29.     else return -1;  
  30. }  
  31. int main()  
  32. {  
  33.     int t,i,j;  
  34.     scanf("%d",&t);  
  35.     while(t--)  
  36.     {  
  37.         scanf("%d%d",&n,&m);  
  38.         for(i=0;i<n;i++)  
  39.         scanf("%d",&s[i]);  
  40.         for(j=0;j<m;j++)  
  41.         scanf("%d",&p[j]);  
  42.         getNext(p);  
  43.         printf("%d\n",kmp(n,m));  
  44.     }  
  45.     return 0;  
  46. }  
#include<stdio.h>
#include<string.h>
int next[10005];
int s[1000005],p[10005];
int n,m;
void getNext(int p[])
{
    int i=0,j=-1,len=m-1;
    next[0]=-1;
    while(i<len)
    {
        if(j==-1||p[i]==p[j])
        {
            i++,j++;
            next[i]=j;}
        else j=next[j];
    }
}
int kmp(int n,int m)
{
    int i=0,j=0;
    while(i<n&&j<m)
    {
        if(j==-1||s[i]==p[j])
        i++,j++;
        else j=next[j];
    }
    if(j==m)return i-j+1;
    else return -1;
}
int main()
{
    int t,i,j;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(i=0;i<n;i++)
        scanf("%d",&s[i]);
        for(j=0;j<m;j++)
        scanf("%d",&p[j]);
        getNext(p);
        printf("%d\n",kmp(n,m));
    }
    return 0;
}


 
posted @ 2014-07-26 08:40  一夜成魔  阅读(145)  评论(0编辑  收藏  举报