HDOJ 1711 Number Sequence(KMP模板题)

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15395    Accepted Submission(s): 6763

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
   
   
   
   

 

 

 

找到子串在母串中出现的位置(子串首字符的位置)。

 

KMP模板题，代码如下：

 

<span style="font-size:12px;">#include<cstdio>
#include<cstring>
#define max 1000010
int T[max],P[10010],next[10010];
int n,m;
void getnext()//获得next数组 
{
	int i=0,j=-1;
	next[0]=-1;
	while(i<m)
	{
		if(j==-1||P[i]==P[j])
		   next[++i]=++j;
		else
		   j=next[j];
	}
}

void kmp()
{
	int i=0,j=0;
	while(i<n)
	{
		if(j==-1||T[i]==P[j])
		{
			i++;
			j++;
			if(j==m)
			{
				printf("%d\n",i-j+1);//字符从零位储存的，注意加1 
				return ;
			}
		}
		else
		   j=next[j];
	}
	printf("-1\n");//母串中找不到子串时，输出-1 
}

int main()
{
	int t,i;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		for(i=0;i<n;++i)
		  scanf("%d",&T[i]);
		for(i=0;i<m;++i)
		  scanf("%d",&P[i]);
		getnext();
		kmp();
	}
	return 0;
}</span>