「杂题乱刷2」CF1454F Array Partition

合集 - 二分(23)

1.「杂题乱刷」AT_abc020_c2023-12-292.「杂题乱刷」CF1846E1 & CF1846E22024-03-053.「杂题乱刷」CF786C2024-04-124.「杂题乱刷」Zheng Rui 327 【2018普转提day17专题】洗2024-05-145.「杂题乱刷」CF1759F2024-05-236.「杂题乱刷」CF1979C2024-06-077.「杂题乱刷」CF1985F2024-06-138.「杂题乱刷」P13962024-06-169.「杂题乱刷」AT_abc360_d2024-07-02

10.「杂题乱刷2」CF1454F Array Partition2024-07-05

11.「杂题乱刷2」CF1996F2024-07-2712.「杂题乱刷2」CF1486C1 & CF1486C22024-08-0713.「杂题乱刷2」CF1360H2024-08-2514.「杂题乱刷2」CF862D2024-08-2915.「杂题乱刷2」CF1301C2024-09-0616.「杂题乱刷2」CF2036G2024-11-0417.「杂题乱刷2」P112672024-11-0918.「杂题乱刷2」CF1370F22024-11-1019.「杂题乱刷2」AT_abc140_e2024-12-1320.「杂题乱刷2」CF601B01-2821.「杂题乱刷2」CF2069D03-0322.「杂题乱刷2」P11830 [省选联考 2025] 幸运数字03-0323.「杂题乱刷2」P11843 [USACO25FEB] The Best Subsequence G03-03

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题目链接

CF1454F Array Partition

解题思路

我们发现显然第一个和第三个区间的值区间随着长度的增大而增大。

于是我们就可以枚举第一个区间的右端点位置，然后现在问题就转化成了找到一个断点来确定第二，三个区间的长度，由于前文提到的第三个区间的值区间随着长度的增大而增大，于是我们就可以通过二分来找出这个断点。

这里我使用了分块来查询区间最小值，时间复杂度 $O(n\times \sqrt{n}\times \log n)$，常数小的话能过。

如果使用其它数据结构可以有更好的复杂度。

参考代码

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#include<bits/stdc++.h>
using namespace std;
//#define map unordered_map
#define re register
#define ll long long
#define forl(i,a,b) for(re ll i=a;i<=b;i++)
#define forr(i,a,b) for(re ll i=a;i>=b;i--)
#define forll(i,a,b,c) for(re ll i=a;i<=b;i+=c)
#define forrr(i,a,b,c) for(re ll i=a;i>=b;i-=c)
#define lc(x) x<<1
#define rc(x) x<<1|1
#define mid ((l+r)>>1)
#define cin(x) scanf("%lld",&x)
#define cout(x) printf("%lld",x)
#define lowbit(x) (x&-x)
#define pb push_back
#define pf push_front
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
#define endl '\n'
#define QwQ return 0;
#define db long double
#define ull unsigned long long
#define lcm(x,y) x/__gcd(x,y)*y
#define Sum(x,y) 1ll*(x+y)*(y-x+1)/2
#define aty cout<<"Yes\n";
#define atn cout<<"No\n";
#define cfy cout<<"YES\n";
#define cfn cout<<"NO\n";
#define xxy cout<<"yes\n";
#define xxn cout<<"no\n";
#define printcf(x) x?cout<<"YES\n":cout<<"NO\n";
#define printat(x) x?cout<<"Yes\n":cout<<"No\n";
#define printxx(x) x?cout<<"yes\n":cout<<"no\n";
#define maxqueue priority_queue<ll>
#define minqueue priority_queue<ll,vector<ll>,greater<ll>>
ll t;
ll n;
ll a[200010];
ll sq,b[200010],maxn[200010],minn[200010],L[200010],R[200010];
ll summax[200010];
ll querymax(ll l,ll r)
{
	ll ma=-1e18;
	if(b[l]==b[r])
	{
		forl(i,l,r)
			ma=max(ma,a[i]);
		return ma;
	}
	forl(i,l,R[b[l]])
		ma=max(ma,a[i]);
	forl(i,L[b[r]],r)
		ma=max(ma,a[i]);
	forl(i,b[l]+1,b[r]-1)
		ma=max(ma,maxn[i]);
	return ma;
}
ll querymin(ll l,ll r)
{
	ll mi=1e18;
	if(b[l]==b[r])
	{
		forl(i,l,r)
			mi=min(mi,a[i]);
		return mi;
	}
	forl(i,l,R[b[l]])
		mi=min(mi,a[i]);
	forl(i,L[b[r]],r)
		mi=min(mi,a[i]);
	forl(i,b[l]+1,b[r]-1)
		mi=min(mi,minn[i]);
	return mi;
}
void solve()
{
	cin>>n;
	forl(i,1,n+1)
		summax[i]=0;
	forl(i,1,n)
		cin>>a[i];
	forr(i,n,1)
		summax[i]=max(summax[i+1],a[i]);
	sq=pow(sqrt(n),1);
	forl(i,1,n)
	{
		L[i]=R[i-1]+1;
		R[i]=min(n,sq*i);
		minn[i]=1e18;
		forl(j,L[i],R[i])
			b[j]=i,minn[i]=min(minn[i],a[j]),maxn[i]=max(maxn[i],a[j]);
		if(R[i]==n)
			break;
	}
	ll ma1=-1e18,ma2=-1e18;
	forl(i,1,n-2)
	{
		ma1=max(ma1,a[i]);
		ll L=i+1,R=n-1;
		while(L<=R)
		{
			ll Mid=(L+R)/2;
			ll mi1=querymin(i+1,Mid);
			ma2=summax[Mid+1];
			if(mi1==ma1 && mi1==ma2)
			{
				cfy;
				cout<<i<<' '<<Mid-i<<' '<<n-Mid<<endl;
				return ;
			}
			if(mi1<=ma1 && ma2<=ma1)
				R=Mid-1;
			else
				L=Mid+1; 
		}
	}
	cfn;
}
int main()
{
	IOS;
	t=1;
 	cin>>t;
	while(t--)
		solve();
	QwQ;
}