Xiamen University Program Design Contest Monthly A (dfs)

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A. CHONG of the ring pigeon (dfs)

Question: Give you a sequence, if all the strings in this sequence are good sequences, then output $chong$ , otherwise output$fuchong$ , the definition of a good sequence is that a unique element can be found in the sequence.

Solution: My previous approach is to enumerate the interval, if the elements in this interval are not unique, output $fuchong$ , but it is obviously timed out,$o(n\ast n)$ complexity, the correct solution (from Niuke) is to find an element that is unique, then if the sequence on the left is a good sequence and the sequence on the right is also a good sequence, then this sequence is a good sequence , This is somewhat similar to the idea of ​​divide and conquer. The specific approach depends on the code.

#include<bits/stdc++.h>
#define mk make_pair
#define pi pair<int,int>
#define pb push_back
#define pp pop_back
using namespace std;

const int maxn = 1e5+100,inf = 1e9+100;
int l[maxn],r[maxn],a[maxn];
map<int,int>mp;

int dfs(int ll,int rr)
{
    
    
	if(ll >= rr)return 1;
	int i = ll ,j = rr;
	while(i <= j)
	{
    
    
		if(l[i] < ll && r[i] > rr)return dfs(ll,i-1)&&dfs(i+1,rr);
		if(l[j] < ll && r[j] > rr)return dfs(ll,j-1)&&dfs(j+1,rr);
		i++,j--;
	}
	return 0;
}

int main()
{
    
    
	int n;
	cin >> n;
	for(int i=1;i<=n;i++)
	{
    
    
		cin >> a[i];
		l[i] = mp[a[i]];
		mp[a[i]] = i;
	}
	mp.clear();
	
	for(int i=n;i>0;i--)
	{
    
    
		if(mp[a[i]]==0)r[i] = inf;
		else r[i] = mp[a[i]];
		mp[a[i]] = i;
	}
	
	int flag = dfs(1,n);
	if(!flag)puts("fuchong");
	else puts("chong");
	return 0;
}