hdu 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

这题是一个最简单的kmp的题目， 子序列的匹配~·

#include<stdio.h>
int bbs,n,m;
int s[1000010];
int a[10010];
int next[10010];
void tonext()  //得到next数组 
{
	int i=0,j=-1;
	next[0]=-1;
	while(i<m)
	{
		if(j==-1||a[i]==a[j])
		{
			i++;j++;
			if(a[i]==a[j])  //如果还是相等  就继续优化 
			{
				next[i]=next[j];
			}
			else next[i]=j;
		}
		else j=next[j];
	}	
}
int kmp()  //序列的匹配 
{ 
	int i=0,j=0;  //2个都设为0 
	while(i<n&&j<m)
	{
		if(a[j]==s[i]||j==-1)  //相等 就比较下一个 
		{
			i++;j++;
		}
		else j=next[j];
	}
	if(j>m-1)  //返回第几个开始 
	return i-m+1;
	else return -1;
}
int main()
{
	scanf("%d",&bbs);
	while(bbs--)
	{
		scanf("%d%d",&n,&m);
		if(n<m)
		{
			printf("-1\n");
			continue;
		}
		for(int i=0;i<n;i++)
		{
			scanf("%d",&s[i]);
		}
		for(int i=0;i<m;i++)
		{
			scanf("%d",&a[i]);
		}		
		tonext();
		int ans=kmp();
		printf("%d\n",ans);
	}
	return 0;
}