hdoj--1171--Number Sequence（KMP）

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 16532    Accepted Submission(s): 7283

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

 

Source

HDU 2007-Spring Programming Contest

#include<stdio.h>
#include<string.h>
int m,n;
int next[1000005];
int a[1000005],b[1000005];
void getnext(int p[])
{
	int j=0,k=-1;
	next[0]=-1;
	while(j<m-1)
	{
		if(k==-1||p[k]==p[j])
		{
			k++;
			j++;
			next[j]=k;
		}
		else k=next[k];
	}
}
int kmp(int a[],int b[])
{
	int i=0,j=0;
	getnext(b);//得到b的next数组
	while(i<n)
	{
		if(j==-1||a[i]==b[j])
		{
			i++;
			j++;
		}
		else
			j=next[j];
		if(j==m)
			return i-m+1;//如果j==m说明b数组已经遍历到底，输出当前a数组元素的下标
	}
	return -1;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int i;
		memset(next,0,sizeof(next));
		scanf("%d%d",&n,&m);
		for(i=0;i<n;i++)
			scanf("%d",&a[i]);
		for(i=0;i<m;i++)
			scanf("%d",&b[i]);
		printf("%d\n",kmp(a,b));
	}
	return 0;
}