F - Number Sequence

KMP算法详解与应用

最新推荐文章于 2020-08-06 10:40:49 发布

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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

题意：输入两组数字:s,t找出第一个s中和t匹配的位置，

思路：KMP模板题。。。。。

下面附上代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1000005;
const int M=10005;
int Next[N],s[N],t[M];
int KMP(int *s,int n,int *t,int m)
{
	int i=0,j=0;
	while(i<n)
	{
		if(j==-1||s[i]==t[j])
		{
			++i,++j;
			if(j==m)
				return i-m+1;
		}
		else 
			j=Next[j];
	}
	return -1;
}
void Getnext(int *t,int l)
{
	int i=0,j=-1;
	Next[i]=-1;
	while(i<l)
	{
		if(j==-1||t[i]==t[j])
			Next[++i]=++j;
		else 
			j=Next[j];
	}
}
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",&s[i]);
		for(int i=0;i<m;i++)
			scanf("%d",&t[i]);
		Getnext(t,m);
		printf("%d\n",KMP(s,n,t,m));
	}
	return 0;
}