KMP（1-模板题）

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
助于理解KMP的链接：http://m.blog.csdn.net/blog/MAOTIANWANG/34466483

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int maxn=1000008;
int t[maxn];
int s[maxn];
int n[maxn];
int len,len1;
void get_next()
{
int i=0;//后移下标
int j=-1;//起始下标
n[0]=-1;
while(i<len)
{
if(j==-1||t[i]==t[j])
{
i++;//若匹配或为起点后移坐标后移
j++;//若匹配或为起点起始坐标后移
n[i]=j;//若匹配则记录前一起始下标情况即匹配数目，不匹配则j=0，记录0
}
else
j=n[j];//如果不匹配则j就倒退回之前达成匹配的最后一位，再次向后匹配，重新来过;
}//当j=0时，j=n[j]=-1；便于进入if使i++，但j依旧是0，没变，让新的i，新的n[i]存储前一j的情况
}
void kmp()
{
int i=0;
int j=0;
while(i<len1)
{
if(s[i]==t[j]||j==-1)
{
i++;
j++;
}
else
j=n[j];
if(j==len)
{
printf("%d\n",i-len+1);
return ;
}
}
printf("-1\n");
}
int main()
{
int m;
scanf("%d",&m);
while(m--)
{
memset(n,0,sizeof(n));
scanf("%d%d",&len1,&len);
for(int i=0; i<len1; i++)
{
scanf("%d",&s[i]);
}
for(int j=0; j<len; j++)
{
scanf("%d",&t[j]);
}
get_next();
kmp();
}
return 0;
}