HDOJ 3480 Division

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      斜率优化一下就能够了。

Division

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 999999/400000 K (Java/Others)
Total Submission(s): 3008    Accepted Submission(s): 1173

Problem Description

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.

 

Input

The input contains multiple test cases.
In the first line of the input there’s an integer T which is the number of test cases. Then the description of T test cases will be given. 
For any test case, the first line contains two integers N (≤ 10,000) and M (≤ 5,000). N is the number of elements in S (may be duplicated). M is the number of subsets that we want to get. In the next line, there will be N integers giving set S.

 

Output

For each test case, output one line containing exactly one integer, the minimal total cost. Take a look at the sample output for format.

 

Sample Input

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

 

Sample Output

Case 1: 1
Case 2: 18


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

 

Source

2010 ACM-ICPC Multi-University Training Contest（5）——Host by BJTU

 

#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
",cas++,dp[m][n]);
    }
	return 0;
}