hdu_1711Number Sequence(kmp)

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 20445    Accepted Submission(s): 8756

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

赤裸裸的kmp

 //kmp
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 int a[],b[];
 int Next[];
 int n,m;
 int kmp()
 {
     int i,j;
     j = ;
     int tm = Next[] = -;
     //ÇóNextÊý×é
     while(j<m-){
         if(tm<||b[j]==b[tm])
             Next[++j] = ++tm;
         else tm = Next[tm];
     }
     //Æ¥Åä
     for( i = j = ; i < n&&j < m; ){
         if(j<||a[i]==b[j])i++,j++;
         else j = Next[j];
     }
     if(j<m) return -;
     return i-j;
 }
 int main()
 {
     int T;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d%d",&n,&m);
         for(int i = ; i< n; i++)
             scanf("%d",&a[i]);
         for(int i = ; i < m; i++)
             scanf("%d",&b[i]);
         int ans = kmp();
         if(ans!=-)
         printf("%d\n",ans+);
         else puts("-1");
     }
     return ;
 }