hdu3480 斜率优化dp

Division

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 999999/400000 K (Java/Others)
Total Submission(s): 2142    Accepted Submission(s): 851

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. 
    
    
    
    
    

      从明显可以得到递推公式dp[i][j]=min(dp[k][j-1]+cos(k+1,i)),dp[i][j]表示第i个数，有j个分组所得的最小和！然后就可以斜率优化了，当然，主要还是要注意状态的转移，倒着写！ 
    
    
    
    
    
 
     
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define MAXN 10005
int prime[MAXN],cut[5050],dp[MAXN][5050],queue[MAXN],s,e;
double ff(int a,int b,int i,int j)
{
    return (((dp[a][j-1]-dp[b][j-1]+prime[a+1]*prime[a+1]-prime[b+1]*prime[b+1])/2.0/(prime[a+1]-prime[b+1])));
}
bool afterm(int a,int b,int c,int i,int j)
{
    if(ff(b,c,i,j)<ff(a,b,i,j))
    return true;
    return false;
}
int fmin(int a,int b)
{

    if(a<b)
        return a;
    return b;
}
bool cmp(int a,int b)
{
    return a<b;
}
int getint()
{
    int sum;
	char c;
    while(c=getchar())
    {
        if(c>='0'&&c<='9')
        {
            sum=c-'0';
			break;
        }
    }
    while(c=getchar())
    {

        if(c<'0'||c>'9')
        {

			break;
        }
		sum=sum*10+c-'0';
    }
    return sum;
}
int main()
{
    int tcase,T,i,n,m,j,k,temp,cnt;
    tcase=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
       for(i=1;i<=n;i++)
        {
        prime[i]=getint();
        }
        if (m >= n)
        {
            printf("0\n");
            continue;
        }
        sort(prime+1,prime+n+1,cmp);
        cnt;
		for (cnt = 0, i = 1; i <= n; i++)
            if (i == 1 || prime[i] != prime[i - 1])
                prime[++cnt] = prime[i];
        n = cnt;
        s=e=0;
        for(i=1;i<=n;i++)
        {
            dp[i][1]=(prime[i]-prime[1])*(prime[i]-prime[1]);
        }
        for(j=2;j<=m;j++)
        {
            {
                s=e=0;
                queue[e++]=j-1;
                for(i=j;i<=n;i++)
                {

                    //temp=queeu;
                    while(s<e-1&&(ff(queue[s],queue[s+1],i,j)<=prime[i]))//¶ÓÍ·×îÓÅ
                    {
                        s++;
                    }
                    k=queue[s];
                    dp[i][j]=dp[k][j-1]+(prime[i]-prime[k+1])*(prime[i]-prime[k+1]);
                    temp=i;
                    while(s<e-1&&afterm(queue[e-2],queue[e-1],temp,i,j))
                    {
                        e--;
                    }
                    queue[e++]=temp;

                }


            }
        }
		printf("Case %d: %d\n",tcase++,dp[n][m]);
    }
    return 0;
}