kmp复习

http://vjudge.net/contest/view.action?cid=47286#problem/A

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>
#include <algorithm>
#include <iostream>
using namespace std;
const int N=1000002;
int a
,next[N/100+2];
int b[N/100+2],m,n;
void get_next()
{
int i,j;
next[0]=j=-1;
for(i=1;i<m;i++)
{
while(j>-1&&b[i]!=b[j+1])
j=next[j];
if(b[i]==b[j+1])
j++;
next[i]=j;
}
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
for(int i=0;i<m;i++)
scanf("%d",&b[i]);
get_next();
int j=-1,ip=0;
int flag=1;
for(int i=0;i<n;i++)
{
while(j>-1&&a[i]!=b[j+1])
{
j=next[j];
}
if(a[i]==b[j+1])
j++;
if(j==m-1)
{
flag=0;
ip=i-j+1;
break;
}
}
if(flag==0)
printf("%d\n",ip);
else
printf("-1\n");
}
return 0;
}