Print Article-斜率优化dp

HDU - 3507 Print Article

\[\begin{align*} dp_i &= \min_{j=1}^{i-1}(dp_j+(sum_i-sum_j)^2+m) \\ dp_i &= dp_j +sum_i^2+sum_j^2-2*sum_i*sum_j+m \\ dp_j+sum_j^2 &= 2*sum_i*sum_j+dp_i-sum_i^2-m\\ y&=k*x+b \end{align*} \]

其中

\[y=dp_j+sum_j^2 \\ k=2*sum_i \\ x=sum_j \\ b=dp_i-sum_i^2-m \]

#include <bits/stdc++.h>
#define int long long
using namespace std;
constexpr int maxn = 5e5+10;
bool Mst;

int n,m;
int q[maxn];
int dp[maxn],sum[maxn];

double slope(int i,int j)
{
    // x-sum[j], y-dp[j]+sum[j]*sum[j]
    // so k:(x_1,y_1),(x_2,y_2)=
    return (double)(dp[i]+sum[i]*sum[i]-dp[j]-sum[j]*sum[j])
        /(sum[i]==sum[j] ? 1e-9 : sum[i]-sum[j]);
    // (y_1-y_2)/(x_1-x_2)---0-无穷大!!!
}

bool Med;
signed main()
{
    // cerr<<1.0*(&Med-&Mst)/1024/1024<<" M\n";

    while(~scanf("%lld%lld",&n,&m))
    {
        for(int i=1;i<=n;++i)
        {
            scanf("%lld",sum+i);
            sum[i]+=sum[i-1];
        }
        int h=1,t=0;
        for(int i=1;i<=n;++i)
        {
            // h<t 保证队列里至少两个点
            // 计算到i时，i-1是i的侯选，尝试加入
            while(h<t && slope(i-1,q[t]) <= slope(q[t],q[t-1]))
            {
                --t;
            }
            q[++t]=i-1;
            while(h<t && slope(q[h+1],q[h])<=2*sum[i])
            {
                ++h;
            }
            int j=q[h];
            dp[i]=dp[j]+(sum[i]-sum[j])*(sum[i]-sum[j])+m;
        }
        printf("%lld\n",dp[n]);
    }

    return 0;
}