Number Sequence (KMP的应用)

个人心得：朴素代码绝对超时，所以要用到KMP算法，特意了解了，还是比较抽象，要多体会

Given two sequences of numbers : a11, a22, ...... , aNN, and b11, b22, ...... , bMM (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make aKK = b11, aK+1K+1 = b22, ...... , aK+M−1K+M−1 = bMM. If there are more than one K exist, output the smallest one. 

InputThe 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 a11, a22, ...... , aNN. The third line contains M integers which indicate b11, b22, ...... , bMM. All integers are in the range of −1000000,1000000−1000000,1000000. 
OutputFor 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

 #include<stdio.h>
 #include<string.h>
 #include<iostream>
 #include<algorithm>
 #include<utility>
 #include<queue>
 #include<set>
 using namespace std;
 int n,m;
 int a[],b[];
 void getnext(int x[],int next[])
 {
     int i=;
     int j=-;
     next[i]=-;
     while(i<m)
     {
         if(j==-||x[i]==x[j])
         {
             i++;
             j++;
             next[i]=j;
         }
         else
             j=next[j];
     }
 }
 int main()
 {
      int t;
      scanf("%d",&t);
      while(t--)
      {
          int next[];
          int flag=-;
          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]);
         getnext(b,next);
         int i=,j=;
           while(i<=n&&j<m)
           {
               if(j==-||a[i]==b[j])
               {
                   i++;
                   j++;
               }
               else
                 j=next[j];
           }
           if(j==m)  flag=i-j;
           cout<<flag<<endl;

      }
     return ;

 }