[APIO2010]特别行动队

解析

转移方程很容易推:\(f_i = \max(f_j + a * (s_i - s_j)^2 + b * (s_i - s_j) + c)\)
然后当 \(j>k\) 时,如果 \(j\) 更优
那么 \(f_j + a * (s_i - s_j)^2 + b * (s_i - s_j) + c > f_k + a * (s_i - s_k)^2 + b * (s_i - s_k) + c\)
整理得:\((f_j + a * s_j^2 - b * s_j) - (f_k + a * s_k^2 - b * s_k) > 2 * a * s_i * (s_j-s_k)\)
因为 \(s_j-s_k\) 大于零
所以我们可以把不等式两边同除 \(s_j-s_k\) (不除 \(2*a\),当然也可以除,但注意 \(a < 0\),除过去要变号)
于是就成了

\[\frac{(f_j + a \times s_j^2 - b \times s_j) - (f_k + a \times s_k^2 - b \times s_k)}{s_j-s_k} > 2 a \times s_i \]

既然是大于号,那么维护上凸壳
右边单调减,单调队列维护即可

\(Code\)

#include<cstdio>
using namespace std;
typedef long long LL;

const int N = 1e6 + 5;
int n , l , r;
LL a , b , c , f[N] , q[N] , s[N];

double slope(int u , int v)
{ 
	return 1.0 * ((f[u] + a * s[u] * s[u] - b * s[u]) - (f[v] + a * s[v] * s[v] - b * s[v])) 
		/ (s[u] - s[v]);
}

int main()
{
	scanf("%d%lld%lld%lld" , &n , &a , &b , &c);
	for(register int i = 1; i <= n; i++) scanf("%lld" , &s[i]) , s[i] += s[i - 1];
	q[l = r = 1] = 0;
	for(register int i = 1; i <= n; i++)
	{
		while (l < r && slope(q[l] , q[l + 1]) > 2.0 * a * s[i]) l++;
		f[i] = f[q[l]] + a * (s[i] - s[q[l]]) * (s[i] - s[q[l]]) + b * (s[i] - s[q[l]]) + c;
		while (r >= l && slope(q[r] , q[r - 1]) < slope(q[r] , i)) r--;
		q[++r] = i;
	}
	printf("%lld" , f[n]);
}
posted @ 2020-09-17 12:51  leiyuanze  阅读(154)  评论(0编辑  收藏  举报