hdu1711

Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8078 Accepted Submission(s): 3670


Problem Description

Given two sequences of numbers : a[1], a[2], ...... , a
, 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
. 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

#include<stdio.h>
#include<string.h>
int next[10005],lena,lenb;
int a[1000005],b[10005];
void set_naxt()//子串的next数组
{
int i=0,j=-1;
next[0]=-1;
while(i<lenb)
{
if(j==-1||b[i]==b[j])
{
i++; j++;
next[i]=j;
}
else
j=next[j];
}
}
int kmp()
{
int i=0,j=0;//比较时j＝0
set_naxt();
while(i<lena)
{
if(j==-1||a[i]==b[j])
{
i++;j++;
}
else
j=next[j];//在这里有可能等于－1，

if(j==lenb)
return i-j+1;
}
return -1;
}
int main()
{
int i,t;
scanf("%d",&t);
while(t--)
{
memset(next,0,sizeof(next));
scanf("%d%d",&lena,&lenb);
for(i=0;i<lena;i++)
scanf("%d",&a[i]);
for(i=0;i<lenb;i++)
scanf("%d",&b[i]);
printf("%d\n",kmp());
}
}