题解 小P的单调数列

传送门

• 当求最大值的题求的是平均值的最大值时：注意两个数求平均值后一定不会比原数中最大的那个更大，所以平均值取得越多越劣

Code:

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 100010
#define ll long long
//#define int long long

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n;
int a[N];

namespace task1{
	double ans;
	ll dp[N], rdp[N];
	void solve() {
		for (int i=n; i; --i) {
			for (int j=i+1; j<=n; ++j) if (a[i]>a[j]) rdp[i]=max(rdp[i], rdp[j]);
			rdp[i]+=a[i];
		}
		for (int i=1; i<=n; ++i) {
			for (int j=1; j<i; ++j) if (a[i]>a[j]) dp[i]=max(dp[i], dp[j]);
			ans=max(ans, 1.0*dp[i]+a[i]);
			ans=max(ans, (dp[i]+rdp[i])/2.0);
			dp[i]+=a[i];
		}
		printf("%.3lf\n", ans);
		exit(0);
	}
}

namespace task2{
	int uni[N], usize, ua[N];
	double ans;
	ll rdp[N];
	struct seg{
		int tl[N<<2], tr[N<<2];
		ll dat[N<<2];
		multiset<ll> s[N<<2];
		#define tl(p) tl[p]
		#define tr(p) tr[p]
		#define dat(p) dat[p]
		#define s(p) s[p]
		#define pushup(p) dat(p)=max(dat(p<<1), dat(p<<1|1))
		void build(int p, int l, int r) {
			tl(p)=l; tr(p)=r;
			if (l==r) return ;
			int mid=(l+r)>>1;
			build(p<<1, l, mid);
			build(p<<1|1, mid+1, r);
		}
		void add(int p, int pos, ll val) {
			if (tl(p)==tr(p)) {s(p).insert(val); dat(p)=*s(p).rbegin(); return ;}
			int mid=(tl(p)+tr(p))>>1;
			if (pos<=mid) add(p<<1, pos, val);
			else add(p<<1|1, pos, val);
			pushup(p);
		}
		void del(int p, int pos, ll val) {
			if (tl(p)==tr(p)) {s(p).erase(val); dat(p)=s(p).empty()?0ll:*s(p).rbegin(); return ;}
			int mid=(tl(p)+tr(p))>>1;
			if (pos<=mid) del(p<<1, pos, val);
			else del(p<<1|1, pos, val);
			pushup(p);
		}
		ll query(int p, int l, int r) {
			if (l<=tl(p) && r>=tr(p)) {return dat(p);}
			int mid=(tl(p)+tr(p))>>1; ll ans=0;
			if (l<=mid) ans=max(ans, query(p<<1, l, r));
			if (r>mid) ans=max(ans, query(p<<1|1, l, r));
			return ans;
		}
	}left, right;
	void solve() {
		for (int i=1; i<=n; ++i) uni[i]=a[i];
		sort(uni+1, uni+n+1);
		usize=unique(uni+1, uni+n+1)-uni-1;
		for (int i=1; i<=n; ++i) ua[i]=lower_bound(uni+1, uni+usize+1, a[i])-uni;
		left.build(1, 1, usize); right.build(1, 1, usize);
		for (int i=n; i; --i) {
			rdp[i]=(ua[i]>1?right.query(1, 1, ua[i]-1):0)+a[i];
			right.add(1, ua[i], rdp[i]);
		}
		// cout<<"rdp: "; for (int i=1; i<=n; ++i) cout<<rdp[i]<<' '; cout<<endl;
		for (int i=1; i<=n; ++i) {
			ll tem=(ua[i]>1?left.query(1, 1, ua[i]-1):0);
			ans=max(ans, 1.0*tem+a[i]);
			ans=max(ans, (tem+rdp[i])/2.0);
			left.add(1, ua[i], tem+a[i]);
			right.del(1, ua[i], rdp[i]);
		}
		printf("%.3lf\n", ans);
		exit(0);
	}
}

signed main()
{
	freopen("b.in", "r", stdin);
	freopen("b.out", "w", stdout);

	n=read();
	for (int i=1; i<=n; ++i) a[i]=read();
	// task1::solve();
	task2::solve();
	
	return 0;
}