[HDU2829]Lawrence

题面描述

传送门

思考

状态转移方程:

$F_{i,p}=\min(F_{j,p-1}+\sum_{x=j+1}^i\sum_{y=x+1}^ia_x*a_y)$

$\sum_{x=j+1}^i\sum_{y=x+1}^ia_x*a_y=\sum_{x=j+1}^ia_x\sum_{y=x+1}^ia_y=\sum_{x=j+1}^ia_x(\sum_{y=1}^ia_y-\sum_{y=1}^xa_y)$

$=(\sum_{x=1}^ia_x-\sum_{x=1}^ja_x)(\sum_{y=1}^ia_y-\sum_{y=1}^xa_y)$ $=(\sum_{x=1}^ia_x-\sum_{x=1}^ja_x)\sum_{y=1}^ia_y-\sum_{x=1}^ia_x\sum_{y=1}^xa_y-\sum_{x=1}^ja_x\sum_{y=1}^xa_y$

设 $s$为 $a$的前缀和，则

$(s_i-s_j)*s_i-\sum_{x=1}^ia_x*s_x+\sum_{x=1}^ja_x*s_x$

设 $sum$为 $\sum_{x=1}a_x*s_x$

则

$(s_i-s_j)*s_i-sum_i+sum_j$

之后随便搞搞得了。

$F_{k,p}-s_i*s_k+sum_k\le F_{j,p}-s_i*s_j+sum_j$

$F_{k,p}-F_{j,p}+sum_k-sum_j\le s_i*(s_k-s_j)(s_k>s_j)$

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;
}