[HDU - 1711] Number Sequence（KMP）

Link：http://acm.hdu.edu.cn/showproblem.php?pid=1711

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 29388    Accepted Submission(s): 12356

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

没什么要注意的，KMP的模板题，找出第二串数与第一串数完全重合时在第一串数中的位置

Code：

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<cstring>
#include<queue>
#include<stack>
using namespace std;
const int maxn=1e6+10;
int a[maxn],b[maxn],n,m;
int p[maxn];
void getnext(int *b,int t)
{
	int i=0,j=0;
	p[0]=-1,j=p[i];
	while(i<t)
	{
		if(j==-1||b[i]==b[j])
			p[++i]=++j;
		else
			j=p[j];
	}
}
int KMP()
{
	int i=0,j=0;
	while(i<n)
	{
		if(j==-1||a[i]==b[j])
		{
			++i,++j;
			if(j==m)
				return i-m+1;
		}
		else
			j=p[j];
	}
	return -1;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		for(int i=0;i<n;i++)
			scanf("%d",&a[i]);
		for(int i=0;i<m;i++)
			scanf("%d",&b[i]);
		getnext(b,m);
		if(n<m)
			printf("-1\n");
		else
			printf("%d\n",KMP());
	}
return 0;
}