【题解】LuoGu2120：[ZJOI2007]仓库建设

原题传送门
初学斜率优化好题
先写出$O(n^2)$DP
预处理$s_i={\sum }_{j=1}^i{x}_j{p}_j,s{p}_i={\sum }_{j=1}^i{p}_j$
转移方程：$dp[i]=min(d{p}_j-s_i+s_j+x_i(s{p}_i-s{p}_j))+c_i$

考虑两个决策$x,y$，满足$x>y$
如果x这个决策比y靠后并且比y更优
那么应该满足$d{p}_x-s_i+s_x+x_i(s{p}_i-s{p}_x)<d{p}_y-s_i+s_y+x_i(s{p}_i-s{p}_y)$
化简得$\frac{(d{p}_x+s_x)-(d{p}_y+s_y)}{s{p}_x-s{p}_y}<x_i$
令$slope(x,y)=\frac{(d{p}_x+s_x)-(d{p}_y+s_y)}{s{p}_x-s{p}_y}$
那么维护单调队列，保证这个相邻两个决策的“斜率”递增
每次取队首更新自己就行了

Code：

#include 
#define maxn 1000010
#define LL long long
using namespace std;
LL n, s[maxn], sp[maxn], c[maxn], dp[maxn], q[maxn], x[maxn];

inline LL read(){
	LL s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}

double slope(int x, int y){ return 1.0 * (dp[x] + s[x] - dp[y] - s[y]) / (sp[x] - sp[y]); }

int main(){
	n = read();
	for (int i = 1; i <= n; ++i){
		x[i] = read(), sp[i] = read(), c[i] = read();
		s[i] = s[i - 1] + x[i] * sp[i], sp[i] += sp[i - 1];
	}
	int h = 0, t = 0;
	for (int i = 1; i <= n; ++i){
		while (h < t && slope(q[h + 1], q[h]) < x[i]) ++h;
		dp[i] = dp[q[h]] - s[i] + s[q[h]] + x[i] * (sp[i] - sp[q[h]]) + c[i];
		while (h < t && slope(q[t], q[t - 1]) > slope(i, q[t])) --t;
		q[++t] = i;
	}
	printf("%lld\n", dp[n]);
	return 0;
}