KMP（1-模板题）

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

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
",i-len+1);
           return ;
       }
   }
   printf("-1
");
}
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;
}