BZOJ 1911

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题目

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数据范围

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做法

$d[i]:=$前$i$个士兵的最大战斗力
$dp[i]=max_{0\leq j <i}(dp[j]+a\times (S[i] - S[j])^2+b\times (S[i]-S[j]) + c)$

显然可以斜率优化

代码

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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAX_N = 1e6 + 5;
int N, deq[MAX_N];
ll x[MAX_N], A, B, C, S[MAX_N], dp[MAX_N];
inline bool check(int x, int y, int z)
{
	ll bx = -2 * A * S[x] + B, dx = A * S[x] * S[x] - B * S[x] + C + dp[x];
	ll by = -2 * A * S[y] + B, dy = A * S[y] * S[y] - B * S[y] + C + dp[y];
	ll bz = -2 * A * S[z] + B, dz = A * S[z] * S[z] - B * S[z] + C + dp[z];
	return (dy - dx) * (bz - by) <= (dz - dy) * (by - bx);
}
inline ll f(int i, int x)
{
	ll tmpS = S[x] - S[i];
	return dp[i] + A * tmpS * tmpS + B * tmpS + C;
}
void solve()
{
	S[0] = 0;
	for (int i = 0; i < N; ++i) S[i + 1] = S[i] + x[i];
	dp[0] = 0;
	deq[0] = 0;
	int s = 0, t = 1;
	for (int i = 1; i <= N; ++i) {
		while (t - s > 1 && f(deq[s], i) <= f(deq[s + 1], i)) s++;
		dp[i] = f(deq[s], i);
		while (t - s > 1 && check(deq[t - 2], deq[t - 1], i)) t--;
		deq[t++] = i;
	}
	printf("%lld\n", dp[N]);
}
int main()
{
	cin >> N >> A >> B >> C;
	for (int i = 0; i < N; ++i) scanf("%lld", &x[i]);
	solve();
	return 0;
}

注意

题目中给了$a<0$的条件，简化了维护凸包的操作。