hduoj 1711 Number Sequence

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

 

这题就是kmp模板题,下面附上代码。

#include<iostream>
typedef long long ll;
using namespace std;
int a[10000],b[1000000];
int Next[10000];//next[x]表示长度为x+1的数列的最大公共前后缀的长度减一（可能有点拗口，但是这样定义是为了kmp方便起见) 
void getnext(int *a,int lena)//获取next数组的值 
{ Next[0]=-1;//一个数的时候一定不存在 
  int k=-1;
  for(int i=1;i<=lena-1;i++)
  { while(k>-1&&a[k+1]!=a[i])//k>-1说明当前子数列还存在最长公共前后缀 ，且不匹配。 
     {k=Next[k];//回溯操作 
	 }
	 if(a[k+1]==a[i]) k++;//匹配的话就加一，这其实是一个递推加上递归的操作，具体的话可以画草图体会一下。 
	 Next[i]=k;
  }
}
int kmp(int *a,int lena,int *b,int lenb)
{ getnext(a,lena);//调用函数获取next数组 
  int k=-1;
  for(int i=0;i<lenb;i++)
  { while(k>-1&&a[k+1]!=b[i])//这里的操作和next一样。 
     {k=Next[k];//回溯操作 
	 }
	 if(a[k+1]==b[i]) k++;
	 if(k==lena-1)//说明已经完全匹配了 
	 { return i-lena+1;
	   //i=i-lena+1;
	   //k=-1;
	 }
  }
  return -1;//不存在 
}
int main()
{  int t,n,m;
   scanf("%d",&t);
   while(t--)
   { scanf("%d%d",&n,&m);
     for(int i=0;i<n;i++) scanf("%d",&b[i]);
     for(int i=0;i<m;i++) scanf("%d",&a[i]);
     int ans=kmp(a,m,b,n);
     printf("%d\n",ans==-1?ans:ans+1);
   }
}

kmp 算法可以说是完美解决了字符串匹配问题，将效率提升到了极限。