[Ybt] [Experimental example 3 of basic calculation deep search class] Insect food calculation

Worm Eater

Topic link: Worm Food Calculation
Thanks $@lzh$ Big guy's help

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Title description

Problem solving ideas

Obviously this is a deep search question (otherwise how could it be included in the deep search column) .

You can search through each letter to find the best answer.

But this is $O(10^{}$The time complexity of${}^n$$)$is definitely not enough, so pruning must be added.

If the current situation is no longer legal, it will naturally not be possible to search.

So now the key is $check\left(\right)$ function.

If all three numbers in the current position have been searched, just judge it directly. But if the number of the previous one is unknown, then at most one can contribute $1$ 。

The last one must be judged specially: whether it is a carry or not, it is illegal.

code

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
using namespace std;

int n,tot,ok;
int v[30];
int b[20];
int ans[30];
char ss[30];
string s[4];

int check()
{
    
    
	int t=0;
	for(int i=n-1;i>=0;i--)
	{
    
    
		int a=ans[s[1][i]-'A'];
		int b=ans[s[2][i]-'A'];
		int c=ans[s[3][i]-'A'];
		if(a!=-1&&b!=-1&&c!=-1)
		{
    
    
			if(t!=-1)
			{
    
    
				if((a+b+t)%n!=c)
					return 0;
				if(i==0&&a+b+t>=n)
					return 0;
				t=(a+b+t)/n;
			}
			else
			{
    
    
				if((a+b)%n!=c&&(a+b+1)%n!=c)
					return 0;
				if(i==0&&a+b>=n)
					return 0;
			}
		}
		else
			t=-1;
	}
	return 1;
}

void dfs(int dep)
{
    
    
	if(ok)
		return;
	if(dep>n)
	{
    
    
		ok=1;
		for(int i=0;i<n;i++)
			cout<<ans[i]<<" ";
		return;
	}
	for(int i=0;i<n;i++)
	{
    
    
		if(!b[i])
		{
    
    
			b[i]=1;
			ans[ss[dep]-'A']=i;
			if(check())
				dfs(dep+1);
			b[i]=0;
			ans[ss[dep]-'A']=-1;
		}
		
	}
}

int main()
{
    
    
	cin>>n;
	cin>>s[1]>>s[2]>>s[3];
	for(int i=n-1;i>=0;i--)
		for(int j=1;j<=3;j++)
			if(!v[s[j][i]-'A'])
			{
    
    
				v[s[j][i]-'A']=1;
				ss[++tot]=s[j][i];
			}
	memset(ans,-1,sizeof(ans));
	dfs(1);
}