Number Sequence ----HDOJ 1711

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9058    Accepted Submission(s): 4149

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

思路：KMP算法，已经略懂

AC代码：

 #include<stdio.h>
 int a[], b[];
 int fail[];
 int n, m, T;
 void getfail()
 {
     fail[] = -;
     int i, j;
     for(i = , j = -; i < m; i ++)
     {
         while(j >=  && b[j + ] != b[i])
         {
             j = fail[j];
         }
         if(b[j + ] == b[i])
             j ++;
         fail[i] = j;
     }
     return ;
 }

 int kmp()
 {
     getfail();
     int i, j;
     for(i = , j = ; i < n;i ++)
     {
         while(j && b[j] != a[i])
         {
             j = fail[j - ] + ;
         }
         if(b[j] == a[i])
             j ++;
         if(j == m)
             return i - m + ;
     }
     return -;
 }

 int main(int argc, char const *argv[])
 {
     scanf("%d", &T);
     int i, j;
     while(T--)
     {
         scanf("%d%d", &n, &m);
         for(i = ; i < n; i ++)
             scanf("%d", &a[i]);
         for(i = ; i < m; i ++)
             scanf("%d", &b[i]);
         getfail();
         printf("%d\n", kmp());
     }
     return ;
 }