斜率优化DP。。。。

对数组排序后，dp【i】【j】表示对前j个物品分i段的最少代价，dp【i】【j】= min｛ dp【i－1】【k】+（a【k+1】－a【j】）^2 }复杂度m*n^2      斜率优化一下就可以了。

Division

Little D is really interested in the theorem of sets recently. There’s a problem that confused him a long time.  
Let T be a set of integers. Let the MIN be the minimum integer in T and MAX be the maximum, then the cost of set T if defined as (MAX – MIN)^2. Now given an integer set S, we want to find out M subsets S1, S2, …, SM of S, such that



and the total cost of each subset is minimal.

2
3 2
1 2 4
4 2
4 7 10 1

Case 1: 1
Case 2: 18


Hint
The answer will fit into a 32-bit signed integer.
 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn=11000;

int n,m;
int dp[maxn/2][maxn],a[maxn];
int q[maxn],head,tail;

int main()
{
	int T_T,cas=1;
	scanf("%d",&T_T);
	while(T_T--)
	{
		scanf("%d%d",&n,&m);
		for(int i=1;i<=n;i++)
			scanf("%d",a+i);
		sort(a+1,a+n+1);
		for(int i=1;i<=n;i++)
			dp[1][i]=(a[i]-a[1])*(a[i]-a[1]);
		for(int i=2;i<=m;i++)
		{
		    head=tail=0;
		    q[tail++]=i-1;
		    for(int j=i;j<=n;j++)
            {
                while(head+1<tail)
                {
                    int p1=q[head];
                    int p2=q[head+1];
                    int x1=a[p1+1],x2=a[p2+1];
                    int y1=dp[i-1][p1]+x1*x1;
                    int y2=dp[i-1][p2]+x2*x2;
                    if((y2-y1)<=(x2-x1)*2*a[j]) head++;
                    else break;
                }
                int k=q[head];
                dp[i][j]=dp[i-1][k]+(a[k+1]-a[j])*(a[k+1]-a[j]);
                while(head+1<tail)
                {
                    int p1=q[tail-2],p2=q[tail-1],p3=j;
                    int x1=a[p1+1],x2=a[p2+1],x3=a[p3+1];
                    int y1=dp[i-1][p1]+x1*x1;
                    int y2=dp[i-1][p2]+x2*x2;
                    int y3=dp[i-1][p3]+x3*x3;
                    if((y3-y2)*(x2-x1)<=(y2-y1)*(x3-x2)) tail--;
                    else break;
                }
                q[tail++]=j;
            }
		}
		printf("Case %d: %d\n",cas++,dp[m][n]);
    }
	return 0;
}

HDOJ 3480 Division,布布扣,bubuko.com

HDOJ 3480 Division

原文：http://blog.csdn.net/ck_boss/article/details/38659509