HDU 1711 Number Sequence (字符串匹配，KMP算法)

HDU 1711 Number Sequence (字符串匹配，KMP算法)

Description

Given two sequences of numbers : a1, a2, ...... , aN, and b1, b2, ...... , bM (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make aK = b1, aK+1 = b2, ...... , aK+M−1 = bM. 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 a1, a2, ...... , aN. The third line contains M integers which indicate b1, b2, ...... , bM. 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

Http

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

Source

字符串匹配，KMP算法

解决思路

就是KMP算法的运用，请参看我的这篇文章。

代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

const int maxN=1000001;
const int inf=2147483647;

int n,m;
int A[maxN];
int B[maxN];
int F[maxN];

int read();

int main()
{
	int TT=read();
	for (int ti=1;ti<=TT;ti++)
	{
		n=read();
		m=read();
		for (int i=0;i<n;i++)
		    A[i]=read();
		for (int i=0;i<m;i++)
		    B[i]=read();
		F[0]=-1;
		for (int i=1;i<m;i++)
		{
			int j=F[i-1];
			while ((B[j+1]!=B[i])&&(j!=-1))
			    j=F[j];
			if (B[j+1]==B[i])
			    F[i]=j+1;
			else
			    F[i]=-1;
		}
		//for (int i=0;i<m;i++)
		//    cout<<F[i]<<' ';
		//cout<<endl;
		int i=0,j=0;
		bool flag=0;
		while (i<n)
		{
			if (A[i]==B[j])
			{
				i++;
				j++;
				if (j==m)
				{
					cout<<i-j+1<<endl;
					flag=1;
					break;
				}
			}
			else
			if (j==0)
			    i++;
			else
			    j=F[j-1]+1;
		}
		if (flag==0)
		    cout<<-1<<endl;
	}
	return 0;
}

int read()
{
	int x=0;
	int k=1;
	char ch=getchar();
	while (((ch<'0')||(ch>'9'))&&(ch!='-'))
	    ch=getchar();
	if (ch=='-')
	{
		k=-1;
		ch=getchar();
	}
	while ((ch<='9')&&(ch>='0'))
	{
		x=x*10+ch-48;
		ch=getchar();
	}
	return x*k;
}