hdoj 1711 Number Sequence 【KMP】

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 15101    Accepted Submission(s): 6623

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
   
   
   
   

 

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
const int N=1000000+10;
const int M=10000+10;
int a[N],b[M];
int next[N];
int ls,lm;
void getn()
{
	int i = 1,j = 0;
	next[0] = -1;
	next[1] = 0;
	while(i < ls)
	{
		if(j == -1||b[i] == b[j])
		{
			i++,j++;
			next[i] = j;  //求出 子串 的模式串 
		}
		else
		j = next[j];
	}
}
int kmp()
{
	
	int i,j;
	getn();
	i = 0;j = 0;
	while(i < lm&&j != ls)
	{
		if(j == -1||a[i]==b[j]) //逐位比较子串与母串 
		{
			i++,j++;
		}
		else
		j = next[j];
	}
	if( j == ls )
		return i-ls+1; //从0开始所以要再加上 1. 
	else
		return -1; //不满足的话 输出 -1. 
}

int main()
{
	int n,i,j,k;
	scanf("%d",&n);
	while(n--)
	{
		scanf("%d%d",&lm,&ls);
		for(i = 0;i < lm; i++)
			scanf("%d",&a[i]);
		for(j = 0;j < ls;j++)
			scanf("%d",&b[j]);
		k = kmp();  
		printf("%d\n",k);		
	}
	return 0;
}