[Ybtoj Chapter 7 Example 4] Word recitation [Hash]

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Problem-solving ideas

Consider passing both the words to be memorized and the words in the article through $Hash$ is transformed into a$hash$ value, and then you can directly find the maximum number of words to be memorized$maxn$ .
For the second question, consider the ruler method:

1. Initialize $l=r=m$ ; (It’s okay to start from 0)
2. Every time $l$ Move to the left, if it is a word to recite, then$num++$
3. If $num==maxn$ , consider whether r can be moved to the left (not the words to be memorized can be moved), and then update the answer$ans=min（ans，r-l)$ Finally cancelrrThe word at position$r$ , see the code for details. .

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Code

#include<iostream>
#include<cstdio>
#include<iomanip>
#include<cstring>
#include<algorithm>
#include<cmath>

using namespace std;

const int p=131,mod=1000007;

int n,m,a[1010],b[100010],v[2000000],use[2000000],vv[2000000],maxn,num,ans=2147483600,r;
string  x;

int hash(string s)
{
    
    
	int cnt=0;
	for(int i=0;i<s.size();i++)
	{
    
    
		cnt=(cnt*p+s[i])%mod;
	}
	return cnt;
}

int main(){
    
    
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
    
    
		cin>>x;
		a[i]=hash(x);
		v[a[i]]=1;
	}	
	scanf("%d",&m);
	for(int i=1;i<=m;i++)
	{
    
    
		cin>>x;
		b[i]=hash(x);
		if(v[b[i]]&&!use[b[i]])
		{
    
    
			maxn++;
			use[b[i]]=1;
		}
	}
	if(maxn==0){
    
    
		printf("0\n0");
		return 0;	
	}
	else printf("%d\n",maxn);
	r=0;
	for(int l=1;l<=m;l++)
	{
    
    
		while(num<maxn&&r<m)
		{
    
    
			r++;
			if(v[b[r]])
			{
    
    	
				if(vv[b[r]]==0)
					num++;
				vv[b[r]]++;
			}
		}
		if(num==maxn)
		{
    
    
			while(!v[b[l]])
				l++;
			ans=min(ans,r-l+1);
			
			if(vv[b[l]]>=1)
			{
    
    
				if(vv[b[l]]==1)
					num--;
				vv[b[l]]--;
			}
		}
	}
	printf("%d",ans);
}