hdu 1711 Number Sequence

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8463    Accepted Submission(s): 3875

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

 

 kmp水

#include<stdio.h>
#include<string.h>
int a[1000005],b[10005],next[10005];
int n,m;
void get_next()
{
     next[0]=-1;
    int i=0,j=-1;
     while(i<m-1)
     {
          if(j==-1||b[i]==b[j])
               next[++i]=++j;
          else
               j=next[j];
     }
}
int kmp()
{
 int i=-1,j=-1;
 while(i<n&&j<m)
 {
       if(j==-1||a[i]==b[j])
          i++,j++;
       else
          j=next[j];
 }
 if(j==m)
     return i-j+1;
 return 0;
}
int main()
{
     int t,i,ans;
     scanf("%d",&t);
     while(t--)
     {
          scanf("%d %d",&n,&m);
          for(i=0;i<n;i++)
               scanf("%d",&a[i]);
          for(i=0;i<m;i++)
               scanf("%d",&b[i]);
               if(n<m)
               {
                    printf("-1\n");
                    continue;
               }
          get_next();
          ans=kmp();
          if(ans)
          {
               printf("%d\n",ans);
          }
          else
               printf("-1\n");

     }
     return 0;
}