Brief description:

略。）

Analysis:

记代价函数 $$!w(x) = ax^2 + bx + c$$

然后：

$$! \begin{aligned} f_i &= \min_{0 \leq j < i}\big\{ f_j + w(s_i-s_j)\big\} \\ &= \min_{0 \leq j < i}\big\{ f_j + as_j^2 - bs_j -2as_is_j\big\} + w(s_i) \end{aligned}$$ 写成斜率优化的标准形式，b = kx + y 这里有：

• $$k = -2as_i$$
• $$x = s_j $$
• $$y = f_j + as_j^2 – bs_j $$

这里 $$x$$ 和 $$k$$ 均单调（因为 $$s_i$$ 单调），单调队列即可。

//}/* .................................................................................................................................. */

const int N = int(1e6) + 9;
LL f[N], s[N], a, b, c; int q[N], cz, op;
int n;

LL det(LL x1, LL y1, LL x2, LL y2){
    return x1*y2 - x2*y1;
}

#define k (-2*a*s[i])
#define x(j) (s[j])
#define y(j) (f[j]+a*sqr(x(j)) - b*(x(j)))
#define eval(j) (k*x(j)+y(j))

int dett(int p0, int p1, int p2){
    LL t = det(x(p1)-x(p0), y(p1)-y(p0), x(p2)-x(p0), y(p2)-y(p0));
    return t < 0 ? -1 : t > 0;
}

LL w(LL x){
    return a*x*x+b*x+c;
}

int main(){

#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif

    RD(n); RDD(a, b, c); REP_1(i, n) s[i] = s[i-1] + RD();

    cz = 0, op = 0; REP_1(i, n){
        while (cz < op && eval(q[cz]) <= eval(q[cz+1])) ++cz; f[i] = eval(q[cz]) + w(s[i]);
        while (cz < op && dett(q[op-1], q[op], i) >= 0) --op; q[++op] = i;
    }

    OT(f[n]);
}

Posted by xiaodao
Category: 日常