hdu 3480 Division 斜率优化

链接：http://acm.hdu.edu.cn/showproblem.php?pid=3480

Division

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 999999/400000 K (Java/Others)
Total Submission(s): 3864    Accepted Submission(s): 1479

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.

题意：把n个数字  分成m个集合。  每个集合的价值是 这个集合中  (max-min)^2。 输出最少的价值

做法：如果想价值最小，不妨先排个序。这样 最后的答案肯定在排序后的数组中 是连续的。

可以得到数组  dp[i][c]=min{dp[j][c-1]+(num[i]-num[j+1])^2

直接dp 复杂度 N*N*M   瞬间爆炸。

可以斜率优化来写；

推导出   

y(j)=dp[j][c-1]+num[j+1]*num[j+1]

x(j)=num[j+1]

当 k<j<i

[y(j)-y(k)]/[x[k]-x[j]]<=2*num[i]

当不等式成立是，j的取值更优。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <malloc.h>
#include <ctype.h>
#include <math.h>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#include <stack>
#include <queue>
#include <vector>
#include <deque>
#include <set>
#include <map>

#define inf 0x7f7f77f
#define ll __int64

int num[10100];
int dp[10100][6010];
int que[10100];

int cc;
int y(int j)
{
    return dp[j][cc-1]+num[j+1]*num[j+1];
}

int main()
{
    int n,m;
    int t;
    scanf("%d",&t);
    int cas=1;
    while(t--)
    {
        scanf("%d%d",&n,&m);//分成m个
        memset(dp,0,sizeof dp);
        
        dp[0][0]=0;
        dp[0][1]=0;
        for(int i=1;i<=n;i++)
        {
            //num[i]=i;
            scanf("%d",num+i);
            
            dp[i][0]=0;
        } 

        sort(num+1,num+1+n);
        for(int i=1;i<=n;i++)
            dp[i][1]=(num[i]-num[1])*(num[i]-num[1]);// 分成0 个 显然不行 0

        for(int c=2;c<=m;c++)// 分成c个
        {
            int tou=0,wei=0;
            que[wei++]=c-1;
            cc=c;
            for(int i=c;i<=n;i++)//长度//  i>=j
            { 
                while(wei-tou>=2)
                {
                    int a=que[tou];
                    int b=que[tou+1];
                    int y1=y(a);
                    int y2=y(b); 
                    int x1=num[a+1];
                    int x2=num[b+1];
                    

                    if((y2-y1)<=2*num[i]*(x2-x1)) 
                        tou++;
                    else
                        break;
                }

                int tem=que[tou];
                dp[i][c]=dp[tem][c-1]+(num[i]-num[tem+1])*(num[i]-num[tem+1]);
            
                while(wei-tou>=2)//g(b,a)>g(c,b)
                {
                    int a=que[wei-2];
                    int b=que[wei-1];
                    int c=i;
                    int y1=y(a);
                    int y2=y(b);
                    int y3=y(c);
                    int x1=num[a+1];
                    int x2=num[b+1];
                    int x3=num[c+1];
                    if((y2-y1)*(x3-x2)>=(x2-x1)*(y3-y2))
                        wei--;
                    else
                        break;
                }
                que[wei++]=i;
            }
            //puts("");
        }
        printf("Case %d: %d\n",cas++,dp[n][m]);
    }
    return 0;
}