[APIO2010]特别行动队

Description

BZOJ1911

Solution

又是一道斜率优化题，有了上一道题的经验，这一道题就很顺利的做完了

关键点：
\(f_i = f_j + a(s_i-s_j)^2+b(si-sj)+c\)
\(j\)比\(k\)优：\(\displaystyle\frac{f_j-f_k+as_j^2-bs_j-as_k^2+bs_k}{2a(s_j-s_k)}<s_i\)
\(F_x = as_x^2 - bs_x + f_x,G_x = 2as_x\)

Code

#include <cstdio>
#include <cmath>
#include <algorithm>

const double eps = 1e-10;
const int N = 1e6+10;
typedef long long LL;

LL a, b, c, n;
LL s[N], t[N], F[N], G[N], f[N];
LL q[N], qhd, qtl;

int dcmp(double x) {
	if (fabs(x) < eps) return 0;
	else if (x > 0) return 1;
	else return -1;
}

double K(int x, int y) {
	return 1.0 * (F[x] - F[y]) / (G[x] - G[y]);
}

bool check(int x, int y, int z) {
	if (dcmp(K(x, y) - K(y, z)) == 1) return true;
	else return false;
}

void dp() {
	q[qtl++] = 0;
	for (int i = 1; i <= n; ++i) {
		while (qhd < qtl-1 && dcmp(K(q[qhd], q[qhd+1]) - s[i]) == -1) qhd++;
		int j = q[qhd];
		f[i] = f[j] + a * (s[i] - s[j]) * (s[i] - s[j]) + b * (s[i] - s[j]) +c;
		F[i] = f[i] + t[i];	
		while (qhd < qtl-1 && check(q[qtl-2], q[qtl-1], i)) qtl--;
		q[qtl++] = i;
	}
}

int main () {
	scanf("%lld%lld%lld%lld", &n, &a, &b, &c);
	LL x;
	for (int i = 1; i <= n; ++i) {
		scanf("%lld", &x);
		s[i] = s[i-1] + x;	
		t[i] = a * s[i] * s[i] - b * s[i];
		G[i] = 2 * a * s[i];
		
	}
	dp();
	printf("%lld\n", f[n]);
	return 0;
}