HDU 1711 Number Sequence (KMP)

最新推荐文章于 2019-07-14 13:24:47 发布

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Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22861 Accepted Submission(s): 9770

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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lcy | We have carefully selected several similar problems for you: 1358 3336 1686 3746 1251

第一次利用kmp写题。。。

刚开始求next函数值时用了next数组，然后就一直CP..

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int n,m;
int a[1000100],b[100100];  
int ne[10010]; 
void getne()
{
	int i,j;
	i=1;j=0;ne[1]=0;
	while(i<m)
	{
		if(j==0||b[i]==b[j])
		{
			++i;++j;
			ne[i]=j;
		}
		else
		j=ne[j];
	}
}
void KMP()
{
	int i,j,ans;
	bool fafe=false; 
	i=1;
	j=1;
	while(i<=n)
	{
        while (j>0&&a[i]!=b[j])  
            j=ne[j];  
        i++;j++;  
        if (j==m+1)  
        {  
            fafe=true;  
            ans=i-m;  
            break;  
        }  
	}
    if (fafe)  
    printf("%d\n",ans);  
    else  
    printf("-1\n"); 
}
int main()
{
	int t,i,j;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		for(i=1;i<=n;i++)
		{
			scanf("%d",&a[i]); 
		}
		for(i=1;i<=m;i++)
		{
			scanf("%d",&b[i]);
		}
		getne();
		KMP();
	}
	return 0;
}