KMP

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11606    Accepted Submission(s): 5294

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 

---

Mean: 

 给你s1,s2两个串，让你找到s2在s1中出现的第一个位置。

analyse:

 KMP字符串水题.

Time complexity：O(n+m)

Source code：

#include<cstdio>
#include<cstring>
int l1,l2;
int a[1000010],b[10010],Next[10010];

void getNext()
{
     Next[0]=0;
     int i,k;
     for(i=1,k=0;i<l2;++i)
     {
           while(b[i]!=b[k] && k>0)
                 k=Next[k-1];
           if(b[i]==b[k]) ++k;
           Next[i]=k;
     }
}
int kmp()
{
     getNext();
     for(int i=0,k=0;i<l1;++i)
     {
           while(a[i]!=b[k] && k>0)
                 k=Next[k-1];
           if(a[i]==b[k]) ++k;
           if(k==l2) return i-l2+2;
     }
     return -1;
}
int main()
{
     int t;
     scanf("%d",&t);
     while(t--)
     {
           scanf("%d %d",&l1,&l2);
           for(int i=0;i<l1;++i) scanf("%d",&a[i]);
           for(int i=0;i<l2;++i) scanf("%d",&b[i]);
           printf("%d ",kmp());
     }
     return 0;
}