HDU1711 Number Sequence

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                                  Number Sequence

                                             Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                   Total Submission(s): 9688    Accepted Submission(s): 4433

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
  
  

 

题解：直接KMP算法。

AC code：

#include <iostream>
#include <string>
using namespace std;
int next[ 10005];
void nextfun(int str1[]){
	int i=1,j=0;
	next[1]=0;
	while(i<=str1[0]){
		if(j==0 || str1[i]==str1[j])
		{
			i++;j++;
			next[i]=j;
		}
		else
			j = next[j];
	}
}
int KMP(int str[],int str1[]){


	int i=1,j=1;
	while(i<=str[0] && j<=str1[0]){

		if(j==0 || str[i]==str1[j]){
			i++;
			j++;
		}
		else
			j = next[j];
	
	}
	if(j>str1[0])
		return i-j+1;
	return -1;

}
int main()
{
//	int str[ 1000005],str1[ 10005];
	int t,n,m,i,j;
	cin>>t;
	while(t--){
		memset(next,0,sizeof(next));
		cin>>n>>m;
		int *str = new int[1000005];
		int *str1 = new int[10005];
		for( j=0;j<n;j++)
			cin>>str[j];
	for(j =0; j<m;j++)
			cin>>str1[j];	
		for(i = n-1; i>=0;i--)
			str[i+1]=str[i];
		str[0]=n;
		for( i = m-1; i>=0;i--)
			str1[i+1]=str1[i];
	       	str1[0]=m;
		nextfun(str1);
     int k = KMP(str,str1);
	 cout<<k<<endl;
		delete [] str;
		delete [] str1;
	}
	return 0;
}