KMP入門題

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 47477    Accepted Submission(s): 19183

 

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

 

 

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難點：注意數的範圍，用字符數組不行，只能用int，而且，用scanf輸入，987ms，用cin超時。

模板題：

#include<iostream>
#include<cstdio>
#include<string.h>
using namespace std;
#define range 1000000
int str[range+5];
int ptr[10005];
void cal_next(int p_len,int *next)
{
	int i,j,k=-1;
	next[0]=-1;
	for(i=1;i<p_len;i++)
	{
		while(k>=0&&ptr[k+1]!=ptr[i])
		k=next[k];
		if(ptr[k+1]==ptr[i])
		k++;
		next[i]=k;
	}
}
int KMP(int s_len,int p_len)
{
	int *next=new int[p_len];
	cal_next(p_len,next);
	
	
	int k=-1;
	for(int i=0;i<s_len;i++)
	{
		while(k>=0&&ptr[k+1]!=str[i])
		k=next[k];
		if(ptr[k+1]==str[i])
		k++;
		if(k==p_len-1)
		{
			cout<<i-p_len+1+1<<endl;
			return 1;
		}
	}
	cout<<-1<<endl;
	return 0;
}
int main()
{
	int n,s_len,p_len;
	cin>>n;
	while(n--)
	{
		
		cin>>s_len>>p_len;
		for(int i=0;i<s_len;i++)
		scanf("%d",&str[i]);
		for(int i=0;i<p_len;i++)
		scanf("%d",&ptr[i]);
		KMP(s_len,p_len);
		
	}
	return 0;
}