[题解]P2120 [ZJOI2007] 仓库建设

思路

我们定义 $dp_i$ 表示运输前 $i$ 个工厂，且在第 $i$ 个位置建立仓库所消耗的最小代价。

那么，我们可以得出状态转移方程：

d p_{i} = min (d p_{j} + x_{i} \times (\sum_{k = j + 1}^{i} p_{k}) - \sum_{k = j + 1}^{i} (x_{k} \times p_{k}) + c_{i})

$dp_i = \min(dp_j + x_i \times (\sum_{k = j + 1}^{i}p_k) - \sum_{k = j + 1}^{i}(x_k \times p_k) + c_i)$

然后用前缀和优化一下：

d p_{i} = min (d p_{j} + x_{i} \times (s p_{i} - s p_{j}) - (s f_{i} - s f_{j}) + c_{i})

$dp_i = \min(dp_j + x_i \times (sp_i - sp_j) - (sf_i - sf_j) + c_i)$

拆开括号得：

d p_{i} = min (d p_{j} - x_{i} \times s p_{j} + s f_{j} + x_{i} \times s p_{i} - s f_{i} + c_{i})

$dp_i = \min(dp_j - x_i \times sp_j + sf_j + x_i \times sp_i - sf_i + c_i)$

如果 $j_2$ 的代价优于 $j_1$ ，那么当且仅当满足以下条件：

d p_{j_{1}} - x_{i} \times s p_{j_{1}} + s f_{j_{1}} + x_{i} \times s p_{i} - s f_{i} + c_{i} > d p_{j_{2}} - x_{i} \times s p_{j_{2}} + s f_{j_{2}} + x_{i} \times s p_{i} - s f_{i} + c_{i}

$dp_{j_1} - x_i \times sp_{j_1} + sf_{j_1} + x_i \times sp_i - sf_i + c_i > dp_{j_2} - x_i \times sp_{j_2} + sf_{j_2} + x_i \times sp_i - sf_i + c_i$

化简得：

d p_{j_{1}} - x_{i} \times s p_{j_{1}} + s f_{j_{1}} > d p_{j_{2}} - x_{i} \times s p_{j_{2}} + s f_{j_{2}}

$dp_{j_1} - x_i \times sp_{j_1} + sf_{j_1} > dp_{j_2} - x_i \times sp_{j_2} + sf_{j_2}$

(d p_{j_{1}} + s f_{j_{1}}) - (d p_{j_{2}} + s f_{j_{2}}) > x_{i} \times (s p_{j_{1}} - s p_{j_{2}})

$(dp_{j_1} + sf_{j_1}) - (dp_{j_2} + sf_{j_2}) > x_i \times (sp_{j_1} - sp_{j_2})$

x_{i} < \frac{(d p_{j_{1}} + s f_{j_{1}}) - (d p_{j_{2}} + s f_{j_{2}})}{s p_{j_{1}} - s p_{j_{2}}}

$x_i < \frac{(dp_{j_1} + sf_{j_1}) - (dp_{j_2} + sf_{j_2})}{sp_{j_1} - sp_{j_2}}$

不妨令：

$dp_{x} + sf_x$ 为 $Y(x)$ 。
$sp_x$ 为 $X(x)$ 。
$x_i$ 为 $K$ 。

那么有：

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

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

然后因为 $p_i$ 有可能为 $0$ ，所以我们找答案时倒着找第一个 $p_i$ 不为 $0$ 的下标 $t$ ，那么，答案就为 $\min_{t \leq i\leq n}{dp_i}$ 。

Code

#include <bits/stdc++.h>  
#define int long long  
#define re register  
  
using namespace std;  
  
const int N = 1e6 + 10,inf = 1e18 + 10;  
int n,hh = 1,tt = 1;  
int x[N],p[N],c[N],sp[N],sf[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 x[i];  
}  
  
inline int X(int i){  
    return sp[i];  
}  
  
inline int Y(int i){  
    return dp[i] + sf[i];  
}  
  
inline double sl(int i,int j){  
    return 1.0 * (Y(j) - Y(i)) / (X(j) - X(i));  
}  
  
signed main(){  
    n = read();  
    for (re int i = 1;i <= n;i++){  
        x[i] = read();  
        p[i] = read();  
        c[i] = read();  
        sp[i] = sp[i - 1] + p[i];  
        sf[i] = sf[i - 1] + x[i] * p[i];  
    }  
    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] + x[i] * (sp[i] - sp[j]) - (sf[i] - sf[j]) + c[i];  
        while (hh < tt && sl(q[tt],q[tt - 1]) >= sl(i,q[tt - 1])) tt--;  
        q[++tt] = i;  
    }  
    int t = n;  
    while (!p[t] && t) t--;  
    if (!t) puts("0");  
    else{  
        int ans = inf;  
        for (re int i = t;i <= n;i++) ans = min(ans,dp[i]);  
        printf("%lld",ans);  
    }  
    return 0;  
}