HDU——1711 Number Sequence

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

在一个串里寻找另一个串第一次出现的位置的下标

kmp的运用。

#include<stdio.h>
#include<string.h>

const int maxn=1000010;

int next[maxn];

int str[maxn],per[maxn];

int len1,len2;

void get_next()
{
	next[0]=-1;
	int j=0,k=-1;
	while(j<len2)
	{
		if(k == -1 || str[k] == str[j])
		{
			j++;
			k++;
			if(str[j] != str[k])
			  next[j]=k;
	        else
	          next[j]=next[k];
		}
		else
		   k=next[k];
	}
}


int KMP()
{
	int index=0;
	int i=0,j=0;
	get_next();
	while(i<len1 && j<len2)
	{
		if(j==-1 || str[j] == per[i])
		{
			i++;
			j++;
		}
		else
			j=next[j];
	}
	if(j==len2)//串匹配到底，返回位置
	  return i-len2+1;
    else //匹配失败，未找到
      return -1;
}
int main()
{
	int t;
	while(~scanf("%d",&t))
	{
		while(t--)
		{
			scanf("%d%d",&len1,&len2);
			for(int i=0;i<len1;i++) 
			   scanf("%d",&per[i]);
            for(int i=0;i<len2;i++) 
			   scanf("%d",&str[i]);
            printf("%d\n",KMP());
		}
	} 
	return 0;
}