【线段树+单调栈】2019牛客暑期多校训练营（第四场）C sequence

链接：https://ac.nowcoder.com/acm/contest/884/C
来源：牛客网
 

时间限制：C/C++ 3秒，其他语言6秒
空间限制：C/C++ 524288K，其他语言1048576K
64bit IO Format: %lld

题目描述

Your are given two sequences a1…na_{1 \dots n}a1…n​ and b1…nb_{1 \dots n}b1…n​ .You need to answer max⁡1≤l≤r≤n{min(al…r)×sum(bl…r)}\displaystyle \max_{1 \le l \le r \le n} \{min(a_{l \dots r}) \times sum(b_{l \dots r})\}1≤l≤r≤nmax​{min(al…r​)×sum(bl…r​)} 。

Where min(a) means the minimal value of every element of sequence a, sum(a) means the sum of every element of sequence a .

输入描述:

The first line contains an integer n .

The second line contains n integers meaning a1…na_{1 \dots n}a1…n​ .

The third line contains n integers meaning b1…nb_{1 \dots n}b1…n​ .

输出描述:

An integer meaning the answer.

示例1

输入

复制

3
1 -1 1
1 2 3

输出

复制

3

备注:

For all test datas, 1≤n≤3×106,−106≤ai,bi≤1061 \leq n \leq 3 \times 10^6,-10^6 \leq a_i,b_i \leq 10^61≤n≤3×106,−106≤ai​,bi​≤106.

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思路：通过以以每个A[i]为最小值所能扩展的区间，然后在此区间内选最大的B数组和，a数组扩展区间可以通过两次单调栈求出，区间和最大可以通过线段树或ST表求出，注意A[i]的正负要分类讨论

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#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
//typedef __int128 bll;
const ll maxn = 3e6+100;
const ll mod = 1e9+7;
const ld pi = acos(-1.0);
const ll inf = 1e18;
const ld eps = 1e-5;
const ld e = exp(1);

ll n,a[maxn],b[maxn],ans,preb[maxn],treema[maxn<<2],treemi[maxn<<2];
struct nodea
{
	ll l,r;
}arra[maxn];

void build(ll now,ll l,ll r)
{
	if(l == r)
	{
		treema[now] = preb[l];
		treemi[now] = preb[l];
		return ;
	}
	
	ll mid = (l+r)/2;
	
	build(now*2,l,mid);
	build(now*2+1,mid+1,r);
	
	treema[now] = max(treema[now*2],treema[now*2+1]);
	treemi[now] = min(treemi[now*2],treemi[now*2+1]);
	return ;
}

ll queryma(ll now,ll l,ll r,ll a,ll b)
{
	if(a <= l && b >= r)
	{
		return treema[now];
	}
	
	ll mid = (l+r)/2;
	
	ll res = -inf;
	
	if(a <= mid)
		res = max( res,queryma(now*2,l,mid,a,b) );
	if(b >= mid+1)
		res = max( res,queryma(now*2+1,mid+1,r,a,b) );
	
	return res;
}

ll querymi(ll now,ll l,ll r,ll a,ll b)
{
	if(a <= l && b >= r)
	{
		return treemi[now];
	}
	
	ll mid = (l+r)/2;
	
	ll res = inf;
	
	if(a <= mid)
		res = min( res,querymi(now*2,l,mid,a,b) );
	if(b >= mid+1)
		res = min( res,querymi(now*2+1,mid+1,r,a,b) );
	
	return res;
}

int main()
{
	
	ios::sync_with_stdio(false);
	cin.tie(0),cout.tie(0);

	cin >> n;
	
	for(ll i = 1; i <= n; i++)
	{
		cin >> a[i];
	}
	
	
	for(ll i = 1; i <= n; i++)
	{
		cin >> b[i];
		preb[i] = preb[i-1] + b[i];
	}
	
	build(1,1,n);
		
	a[0] = -inf;
	stack<ll>S;
	S.push(0);
	for(ll i = 1; i <= n; i++)
	{
		while(a[S.top()] >= a[i])
		{
			S.pop();	
		}	
		arra[i].l = S.top();
		S.push(i);
	}
	
	stack<ll>SS;
	a[n+1] = -inf;
	SS.push(n+1);
	for(ll i = n; i >= 1; i--)
	{
		while(a[SS.top()] >= a[i])
		{
			SS.pop();
		}
		arra[i].r = SS.top();
		SS.push(i);
	}
	
	for(ll i = 1; i <= n; i++)
	{
		ll l = arra[i].l,r = arra[i].r,num = a[i];
				
		if(a[i] > 0)
		{
			ll tr = queryma(1,1,n,i,r-1);
			ll tl = querymi(1,1,n,l,i);
			ans = max(ans,1ll*a[i]*(tr-tl));			
		}
		else
		{
			ll tr = querymi(1,1,n,i,r-1);
			ll tl = queryma(1,1,n,l,i);
			ans = max(ans,1ll*a[i]*(tr-tl));
			
		}
		
	}
	
	cout << ans << endl;
	
	return 0;	
}