2018.09.08 bzoj4518: [Sdoi2016]征途（斜率优化dp）

传送门

把式子展开后发现就是要求：

m∗(∑i=1msum′[i])−sum[n]2" role="presentation" style="position: relative;">m∗(∑mi=1sum′[i])−sum[n]2m∗(∑i=1msum′[i])−sum[n]2的最小值。

于是只需要求：

m∗(∑i=1msum′[i])" role="presentation" style="position: relative;">m∗(∑mi=1sum′[i])m∗(∑i=1msum′[i])的最小值。

于是设f[i][j]" role="presentation" style="position: relative;">f[i][j]f[i][j]表示前i个分了j组的最小值。

显然有：

f[i][j]=min(f[k][j−1]+(sum[i]−sum[k])2)" role="presentation" style="position: relative;">f[i][j]=min(f[k][j−1]+(sum[i]−sum[k])2)f[i][j]=min(f[k][j−1]+(sum[i]−sum[k])2)

<=>

f[i][j]=min(f[k][j−1]+sum[k]2−2sum[i]∗sum[k])+sum[i]2" role="presentation" style="position: relative;">f[i][j]=min(f[k][j−1]+sum[k]2−2sum[i]∗sum[k])+sum[i]2f[i][j]=min(f[k][j−1]+sum[k]2−2sum[i]∗sum[k])+sum[i]2

对于两个决策k1&lt;k2" role="presentation" style="position: relative;">k1<k2k1<k2且k2比k1更优，有：

f[k1][j−1]+sum[k1]2−2sum[i]∗sum[k1]" role="presentation" style="position: relative;">f[k1][j−1]+sum[k1]2−2sum[i]∗sum[k1]f[k1][j−1]+sum[k1]2−2sum[i]∗sum[k1]

>

f[k2][j−1]+sum[k2]2−2sum[i]∗sum[k2]" role="presentation" style="position: relative;">f[k2][j−1]+sum[k2]2−2sum[i]∗sum[k2]f[k2][j−1]+sum[k2]2−2sum[i]∗sum[k2]

令t[k]=f[k][j−1]+sum[k]2" role="presentation" style="position: relative;">t[k]=f[k][j−1]+sum[k]2t[k]=f[k][j−1]+sum[k]2

=>

(t[k1]−t[k2])/(sum[k1]−sum[k2])&lt;2sum[i]" role="presentation" style="position: relative;">(t[k1]−t[k2])/(sum[k1]−sum[k2])<2sum[i](t[k1]−t[k2])/(sum[k1]−sum[k2])<2sum[i]

果断斜率优化。

代码：

#include<bits/stdc++.h>
#define ll long long
#define N 3005
using namespace std;
inline ll read(){
    ll ans=0;
    char ch=getchar();
    while(!isdigit(ch))ch=getchar();
    while(isdigit(ch))ans=(ans<<3)+(ans<<1)+(ch^48),ch=getchar();
    return ans;
}
int n,m,hd,tl,q[N];
ll sum[N],f[N][N];
inline double slope(int k,int i,int j){return 1.0*(f[i][k]+sum[i]*sum[i]-f[j][k]-sum[j]*sum[j])/(sum[i]-sum[j]);}
int main(){
    n=read(),m=read();
    for(int i=1;i<=n;++i)sum[i]=sum[i-1]+read(),f[i][0]=1e18;
    for(int j=1;j<=m;++j){
        hd=tl=1,q[1]=0;
        for(int i=1;i<=n;++i){
            while(hd<tl&&slope(j-1,q[hd+1],q[hd])<2.0*sum[i])++hd;
            int k=q[hd];
            f[i][j]=f[k][j-1]+(sum[i]-sum[k])*(sum[i]-sum[k]);
            while(hd<tl&&slope(j-1,q[tl],q[tl-1])>slope(j-1,i,q[tl]))--tl;
            q[++tl]=i;
        }
    }
    cout<<(m*f[n][m]-sum[n]*sum[n]);
    return 0;
}