AcWing 构造数组(单调栈)

Posted 2021-08-26 爷灬傲奈我何123

构造数组
题意：

思路：
一开始想的枚举每一个山峰，然后往两边取min，然后TLE不知道怎么优化，正解是考虑单调栈，每一个l，r数组维护前缀、后缀的最大匹配值，例如l数组，对于a[i]而言，左边和它相邻所有比他大的数的贡献都是a[i]直到找到第一个比他小的数,那么就是a[i]*长度+l[stk[top]];

// Problem: 构造数组
// Contest: AcWing
// URL: https://www.acwing.com/problem/content/description/3783/
// Memory Limit: 256 MB
// Time Limit: 3000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define x first
#define y second
typedef pair<int,int> pii;
const int N=500010;
ll l[N];
ll r[N];
int a[N];
int stk[N];
int ans[N];
int main()
{
	int n;
	cin >> n;
	for(int i=1;i<=n;i++)
	{
		cin >> a[i];
	}
	int top= 0;
	for(int i=1;i<=n;i++)
	{
		if(!top)	
		{
			stk[++top]=i;
			l[i]=1ll*i*a[i];
		}
		else
		{
			while(top && a[i]<=a[stk[top]])	top--;
			if(!top)
			{
				l[i]=1ll*i*a[i];
			}
			else{
				l[i]=l[stk[top]]+1ll*(i-stk[top])*a[i];
			}
			stk[++top]=i;
		}
	}
	top=0;
	for(int i=n;i>=1;i--)
	{
		if(!top)
		{
			stk[++top]=i;
			r[i]=1ll*(n-i+1)*a[i];
		}
		else
		{
			while(top && a[i]<=a[stk[top]])	top--;
			if(!top)
			{
				r[i]=1ll*(n-i+1)*a[i];
			}
			else
			{
				r[i]=r[stk[top]]+1ll*(stk[top]-i)*a[i];
			}
			stk[++top]=i;
		}
	}
	ll res=0;
	int id=0;
	for(int i=1;i<=n;i++)
	{
		if(l[i]+r[i]-a[i]>res)
		{
			id=i;
			res=l[i]+r[i]-a[i];
		}
	}
	for(int i=id;i<=n;i++)
	{
		if(i==id)	ans[i]=a[i];
		else
		{
			ans[i]=min(a[i],ans[i-1]);
		}
	}
	for(int i=id-1;i>=1;i--)
	{
		ans[i]=min(ans[i+1],a[i]);
	}
	for(int i=1;i<=n;i++)
	cout<< ans[i] <<" ";
	cout<<endl;
	
}
/**
* In every life we have some trouble
* When you worry you make it double
* Don't worry,be happy.
**/