P3195 [HNOI2008]玩具装箱

先推出转移方程
$f_{i} = min_{j < i} {f_{j} + (s_{i} - s_{j} + i - j - 1 - L)^{2}}$
为了方便，令
$s_{i} = s_{i} + i$

$L = L + 1$

那么方程为
$f_{i} = min {f_{j} + (s_{i} - s_{j} - L)^{2}}$
化为斜率优化的形式
$f_{i} - (s_{i} - L)^{2} = min_{j < i} {f_{j} + s_{j}^{2} + 2 s_{j} (L - s_{i})}$
因为 $s_{i}$ 单增， $s_{i} + i$ 也是单增，用单调队列维护这个凸包即可。

查看代码

#include <cstdio>
using namespace std;
typedef long long LL;
const int N = 5e4 + 10, inf = 1e9;
int hd = 1, tl, q[N];
int n, L, w[N];
LL s[N], f[N];
double slope(int a, int b)
{
    return (double)(f[a] + s[a] * s[a] - f[b] - s[b] * s[b]) / (double)(s[a] - s[b]);
}
int main()
{
    scanf("%d%d", &n, &L);
    for (int i = 1; i <= n; i++)
        scanf("%d", &w[i]);
    for (int i = 1; i <= n; i++)
        s[i] = s[i - 1] + w[i];
    for (int i = 1; i <= n; i++)
        s[i] += i;
    L++;
    q[++tl] = 0;
    for (int i = 1; i <= n; i++)
    {
        while (hd < tl && slope(q[hd], q[hd + 1]) < 2 * (s[i] - L))
            hd++;
        f[i] = f[q[hd]] + (s[i] - s[q[hd]] - L) * (s[i] - s[q[hd]] - L);
        while (hd < tl && slope(q[tl], q[tl - 1]) > slope(i, q[tl]))
            tl--;
        q[++tl] = i;
    }
    printf("%lld", f[n]);
    return 0;
}