Number Sequence

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

解题思路

这个也就是KMP的板子问题吧，比较简单只用把字符换成数字就行了

代码

#include<string>
#include<vector>
#include<iostream>
#include<string>

using namespace std;
#define vec vector<int>
int n,m,N;
int v[6000000];
vec nex(int  ne[])
{
	vec net(N);
	int j;
	for(int i=1;i<N;i++)
	{
		j=net[i-1];
		while(ne[i]!=ne[j]&&j>0)
		{
			j=net[j-1];
		}
		if(ne[i]==ne[j]) j++;
		net[i]=j;
	}
	return net;
}

int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		N=n+m+1;
		for(int i=m+1;i<N;i++) scanf("%d",&v[i]);
		for(int i=0;i<m;i++) scanf("%d",&v[i]);
		v[m]=1000009;
		int flag=-1;
		vec V=nex(v);
		for(int i=m+1;i<N;i++)
		{
			if(V[i]==m)
			{
				flag=i-m-m+1;
				break;
			}
		}
		printf("%d\n",flag);
	}	
	return 0;
}