[题解]P3628 [APIO2010] 特别行动队

思路

我们定义 $dp_i$ 为选取前 $i$ 个人所得到的最大的价值。

那么，我们能得出一个很简单的状态转移方程：

d p_{i} = max (d p_{j} + a \times (\sum_{k = j + 1}^{i} x_{i})^{2} + b \times (\sum_{k = j + 1}^{i} x_{i}) + c)

$dp_i = \max(dp_j + a \times (\sum_{k = j + 1}^{i}x_i)^2 + b \times (\sum_{k = j + 1}^{i}x_i) + c)$

用前缀和优化，得：

d p_{i} = max (d p_{j} + a \times (s x_{i} - s x_{j})^{2} + b \times (s x_{i} - s x_{j}) + c)

$dp_i = \max(dp_j + a \times (sx_i - sx_{j})^2 + b \times (sx_i - sx_j) + c)$

然后，拆开括号，得：

d p_{i} = max (d p_{j} + a \times s x_{i}^{2} - 2 a \times s x_{i} \times s x_{j} + a \times s x_{j}^{2} + b \times s x_{i} - b \times s x_{j} + c)

$dp_i = \max(dp_j + a \times sx_i^2 - 2a \times sx_i \times sx_j + a \times sx_j^2 + b \times sx_i - b \times sx_j + c)$

如果当前选 $j_2$ 优于 $j_1$ ，那么，当且仅当满足如下条件：

d p_{j_{1}} + a \times s x_{i}^{2} - 2 a \times s x_{i} \times s x_{j_{1}} + a \times s x_{j_{1}}^{2} + b \times s x_{i} - b \times s x_{j_{1}} + c < d p_{j_{2}} + a \times s x_{i}^{2} - 2 a \times s x_{i} \times s x_{j_{2}} + a \times s x_{j_{2}}^{2} + b \times s x_{i} - b \times s x_{j_{2}} + c

$dp_{j_1} + a \times sx_i^2 - 2a \times sx_i \times sx_{j_1} + a \times sx_{j_1}^2 + b \times sx_i - b \times sx_{j_1} + c <\\ dp_{j_2} + a \times sx_i^2 - 2a \times sx_i \times sx_{j_2} + a \times sx_{j_2}^2 + b \times sx_i - b \times sx_{j_2} + c$

化简得：

2 a \times s x_{i} < \frac{(d p_{j_{2}} + a \times d p_{j_{2}}^{2} - b \times s x_{j_{2}}) - (d p_{j_{1}} + a \times d p_{j_{1}}^{2} - b \times s x_{j_{1}})}{s x_{j_{2}} - s x_{j_{2}}}

$2a \times sx_i < \frac{(dp_{j_2} + a \times dp_{j_2}^2 - b \times sx_{j_2}) - (dp_{j_1} + a \times dp_{j_1}^2 - b \times sx_{j_1})}{sx_{j_2} - sx_{j_2}}$

令：

$2a \times sx_i$ 为 $K$ 。
$sx_x$ 为 $X(x)$ 。
$dp_x + a \times sx_x^2 - b \times sx_x$ 为 $Y(x)$ 。

那么，得：

K < \frac{Y (j_{2}) - Y (j_{1})}{X (j_{2}) - X (j_{1})}

$K < \frac{Y(j_2) - Y(j_1)}{X(j_2) - X(j_1)}$

那么，要使答案尽可能的大，最有抉择点必定在上凸包上，用斜率优化 DP 维护即可。

Code

#include <bits/stdc++.h>  
#define int long long  
#define re register  
  
using namespace std;  
  
const int N = 1e6 + 10;  
int n,a,b,c,hh = 1,tt = 1;  
int arr[N],dp[N],q[N];  
  
inline int read(){  
    int r = 0,w = 1;  
    char c = getchar();  
    while (c < '0' || c > '9'){  
        if (c == '-') w = -1;  
        c = getchar();  
    }  
    while (c >= '0' && c <= '9'){  
        r = (r << 3) + (r << 1) + (c ^ 48);  
        c = getchar();  
    }  
    return r * w;  
}  
  
inline int K(int i){  
    return 2 * a * arr[i];  
}  
  
inline int X(int i){  
    return arr[i];  
}  
  
inline int Y(int i){  
    return dp[i] + a * arr[i] * arr[i] - b * arr[i];  
}  
  
inline double sl(int i,int j){  
    return 1.0 * (Y(i) - Y(j)) / (X(i) - X(j));  
}  
  
signed main(){  
    n = read();  
    a = read();  
    b = read();  
    c = read();  
    for (re int i = 1;i <= n;i++) arr[i] = arr[i - 1] + read();  
    for (re int i = 1;i <= n;i++){  
        while (hh < tt && K(i) <= sl(q[hh],q[hh + 1])) hh++;  
        int j = q[hh];  
        dp[i] = dp[j] + a * (arr[i] - arr[j]) * (arr[i] - arr[j]) + b * (arr[i] - arr[j]) + c;  
        while (hh < tt && sl(q[tt],q[tt - 1]) <= sl(i,q[tt - 1])) tt--;  
        q[++tt] = i;  
    }  
    printf("%lld",dp[n]);  
    return 0;  
}