題目連結：http://acm.hdu.edu.cn/showproblem.php?pid=1711

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11475    Accepted Submission(s): 5234


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
Source
思路：KMP的第一種寫法：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
const int N=1000100;
using namespace std;
int a[N],b[N],next[N];
int n,m;

void get_next()
{
  int i=1,j=0;
  next[1]=0;
  while(i<m)
  {
    if(j==0||b[i]==b[j])
    {
      ++i;
      ++j;
      next[i]=j;
    }
    else
      j=next[j];
  }
}

int kmp()
{
 int i=1,j=1;
 while(i<=n&&j<=m)
 {
  if(j==0||a[i]==b[j])
  {
     ++i;
     ++j;
  }
  else
    j=next[j];
 }
 if(j>m)return i-m;
return -1;
}

int main()
{
    int T;
    cin>>T;
    while(T--)
    {
     scanf("%d%d",&n,&m);
     memset(a,0,sizeof(a));
     memset(b,0,sizeof(b));
     memset(next,0,sizeof(next));
     for(int i=1;i<=n;i++)scanf("%d",&a[i]);
     for(int i=1;i<=m;i++)scanf("%d",&b[i]);
     get_next();
     printf("%d\n",kmp());
    }
    return 0;
}
 

KMP的第二種寫法：

#include <cstdlib>
#include <iostream>
#include <stdio.h>
#include <string>
#include <string.h>
using namespace std;
#define N 1000000+10
int len,len1,next[N];
int str[N],str1[N];

void get_next()
{
   int i=0,j=-1;
   next[0]=-1;
   while(i<=len1)
   {
    if(j==-1||str1[i]==str1[j])
    {
    i++;j++;
    next[i]=j;
    }
    else
      j=next[j];
   }
}

int kmp()
 {
 int i=0,j=0;
 while(i<len&&j<len1)
 {
  if(j==-1||str[i]==str1[j])
  {
   i++;j++;
  }
  else
  j=next[j];
}
  if(j>=len1)
    return i-len1+1;
  return 0;
}

int main(int argc, char *argv[])
{ int kk;
   scanf("%d",&kk);
   while(kk--){
     memset(str,0,sizeof(str));
     memset(str1,0,sizeof(str1));
     memset(next ,0,sizeof(next));
     scanf("%d%d",&len,&len1);
    for(int i=0;i<len;++i)
        scanf("%d",&str[i]);
      for(int i=0;i<len1;++i)
          scanf("%d",&str1[i]);
     get_next();
     int flag=kmp();
    if(flag==0)
         printf("-1\n");
     else
          printf("%d\n",flag);
    }

   return 0;
}