题面描述
传送门
思考
状态转移方程:
Fi,p=min(Fj,p−1+x=j+1∑iy=x+1∑iax∗ay)
x=j+1∑iy=x+1∑iax∗ay=x=j+1∑iaxy=x+1∑iay=x=j+1∑iax(y=1∑iay−y=1∑xay)
=(x=1∑iax−x=1∑jax)(y=1∑iay−y=1∑xay)
=(x=1∑iax−x=1∑jax)y=1∑iay−x=1∑iaxy=1∑xay−x=1∑jaxy=1∑xay
设
s为
a的前缀和,则
(si−sj)∗si−x=1∑iax∗sx+x=1∑jax∗sx
设
sum为
∑x=1ax∗sx
则
(si−sj)∗si−sumi+sumj
之后随便搞搞得了。
Fk,p−si∗sk+sumk≤Fj,p−si∗sj+sumj
Fk,p−Fj,p+sumk−sumj≤si∗(sk−sj)(sk>sj)
AC code
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<cstring>
#define ll long long
#define gc getchar()
using namespace std;
const int N=1e3+10;
inline void qr(ll &x)
{
x=0;char c=gc;int f=1;
while(c<'0'||c>'9'){if(c=='-')f=-1;c=gc;}
while(c>='0'&&c<='9'){x=x*10+(c^48);c=gc;}
x*=f;
}
void qw(ll x)
{
if(x<0)x=-x,putchar('-');
if(x/10)qw(x/10);
putchar(x%10+48);
}
ll s[N],sum[N],f[N],g[N],a[N];
int q[N],l,r,n,m;
inline ll calc1(int j,int k)
{
return g[k]-g[j]+sum[k]-sum[j];
}
inline ll calc2(int j,int k)
{
return s[k]-s[j];
}
bool pd(int i,int j,int k)
{
return calc1(j,i)*calc2(k,j)<=calc1(k,j)*calc2(j,i);
}
int main()
{
while(~scanf("%d%d",&n,&m)&&n&&m)
{
memset(s,0,sizeof(s));memset(sum,0,sizeof(sum));m++;
for(int i=1;i<=n;i++)qr(a[i]),s[i]=s[i-1]+a[i],sum[i]=sum[i-1]+a[i]*s[i];
for(int i=1;i<=n;i++)f[i]=s[i]*s[i]-sum[i];
for(int j=2;j<=m;j++)
{
memcpy(g,f,n<<3);l=1;r=1;q[1]=0;
for(int i=1;i<=n;i++)
{
while(l<r&&calc1(q[l],q[l+1])<=s[i]*calc2(q[l],q[l+1]))++l;
while(l<r&&pd(i,q[r],q[r-1]))--r;
q[++r]=i;f[i]=g[q[l]]+(s[i]-s[q[l]])*s[i]-sum[i]+sum[q[l]];
}
}
qw(f[n]);puts("");
}
return 0;
}