HDU - 1711 Number Sequence KMP

https://vjudge.net/problem/HDU-1711

HDU - 1711 Number Sequence

• • 题目
  • AC代码

题目

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

kmp模板题

AC代码

#include<iostream>
using namespace std;
const int N=1e6+10,M=1e4+10;
int n,m;
int ne[M];
int s[N],p[M];
int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		cin>>n>>m;
		for(int i=0;i<n;i++)	scanf("%d",&s[i]);
		for(int i=0;i<m;i++)	scanf("%d",&p[i]);
		
		ne[0]=-1;
		for(int i=0,j=-1;i<m;)
		{
			if(j==-1 || p[i]==p[j])
			{
				i++,j++;
				ne[i]=j;
			}
			else
				j=ne[j];
		}
		
		int f=0;
		for(int i=0,j=0;i<n;)
		{
			if(j==-1 || s[i]==p[j])
			{
				i++;j++;
			}
			else
				j=ne[j];
			if(j==m)
			{
				f=1;
				printf("%d\n",i-m+1);
				break;
			}
		}
		if(!f)
			printf("-1\n");
	}
	return 0;
}