HDU 3480 Division（斜率优化+二维DP）

　　　　　　　　　　　　　　　　Division

　　　　　　　　　　　　　　　　　　　　　　Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 999999/400000 K (Java/Others)
　　　　　　　　　　　　　　　　　　　　　　　　　　　　　Total Submission(s): 3984    Accepted Submission(s): 1527

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

 

【思路】

       斜率优化+分配式DP。

       设f[i][j]表示将前i个分作j个集合所得最小消费，则有转移方程式：

             f[i][j]=min{ f[k][j-1]+(A[k]-A[j+1])^2 }   

       若有k>l，且决策k优于决策l则有：

              f[k][j-1]-f[l][j-1]+sq(A[k+1])-sq(A[l+1]) <= 2*(A[k+1]-A[l+1])*A[i]

       先进行j循环枚举f[][j]，每一层维护一个单调队列即可。

       乘除耗费时间悬殊，如果直接除这个题就超时了。

【代码】

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;

typedef double Do;
const int N = 1e4+10;
const int M = 5000+10;

int f[N][M],A[N],q[N];
int n,m,L,R;
int sq(int x) { return x*x; }
int UP(int l,int k,int j) {
    return f[k][j-1]-f[l][j-1]+sq(A[k+1])-sq(A[l+1]);
}
int DN(int l,int k,int j) {
    return 2*(A[k+1]-A[l+1]);
}
void read(int& x) {
    char c=getchar(); while(!isdigit(c)) c=getchar();
    x=0; while(isdigit(c)) x=x*10+c-'0' , c=getchar();
}
int main() {
    int T,kase=0;
    read(T);
    while(T--) {
        read(n),read(m);
        for(int i=1;i<=n;i++) read(A[i]);
        sort(A+1,A+n+1);
        for(int i=1;i<=n;i++) f[i][1]=sq(A[i]-A[1]);    //初始化第一层 
        for(int j=2;j<=m;j++) {
            L=R=0;
            for(int i=1;i<=n;i++) {
                while(L<R && UP(q[L],q[L+1],j)<=A[i]*DN(q[L],q[L+1],j)) L++;
                int t=q[L];
                f[i][j]=f[t][j-1]+sq(A[i]-A[t+1]);
                while(L<R && UP(q[R-1],q[R],j)*DN(q[R],i,j)>=UP(q[R],i,j)*DN(q[R-1],q[R],j)) R--;
                q[++R]=i;
            }
        }
        printf("Case %d: %d
",++kase,f[n][m]);
    }
    return 0;
}