HDU3480 Division —— 斜率优化DP

题目链接：https://vjudge.net/problem/HDU-3480

 

Division

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 999999/400000 K (Java/Others)
Total Submission(s): 5304    Accepted Submission(s): 2093

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

 

 

Recommend

zhengfeng

 

 

题意：

给出一组数，把这组数分成m个集合，使得每个集合的（MAX-MIN）^2的和最小。

 

题解：

1.首先可以确定：每个集合的数值跨度应该尽量小，所以可以先对这些数进行排序，被分成一组的数必定是相连的。

2.设dp[i][j]为：第j个数属于第i个集合时的最小值，那么：dp[i][j] = min（dp[i-1][k] + （val[i] - val[k+1）^2），其中 i-1<=k<=j-1。

3.根据上述的状态转移方程，可算得时间复杂度为O（n^3），无法接受。因此可以用斜率优化。

 

 

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int mod = 1e9+7;
const int MAXM = 1e5+10;
const int MAXN = 1e4+10;

int val[MAXN], dp[MAXN][MAXN];
int q[MAXN], head, tail;

int getUP(int i, int k1, int k2)
{
    return (dp[i-1][k1] + val[k1+1]*val[k1+1])-
            (dp[i-1][k2] + val[k2+1]*val[k2+1]);
}

int getDOWN(int k1, int k2)
{
    return 2*(val[k1+1]-val[k2+1]);
}

int getDP(int i, int j, int k)
{
    return dp[i-1][k] + (val[j]-val[k+1])*(val[j]-val[k+1]);
}

int main()
{
    int n, m, T;
    scanf("%d", &T);
    for(int kase = 1; kase<=T; kase++)
    {
        scanf("%d%d", &n,&m);
        for(int i = 1; i<=n; i++)
            scanf("%d", &val[i]);

        sort(val+1, val+1+n);
        for(int i = 1; i<=n; i++)   //初始化第一段
            dp[1][i] = (val[i]-val[1])*(val[i]-val[1]);
        for(int i = 2; i<=m; i++)   //从i-1段转移到i段
        {
            head = tail = 0;
            q[tail++] = i-1;   //i-1段最少要有i-1个数，故从i-1开始 
            for(int j = i; j<=n; j++)   //i段最少要有i个数，故从i开始
            {
                while(head+1<tail && getUP(i,q[head+1],q[head])<getDOWN(q[head+1], q[head])*val[j]) head++;
                dp[i][j] = getDP(i,j,q[head]);

                while(head+1<tail && getUP(i,j,q[tail-1])*getDOWN(q[tail-1],q[tail-2])<=
                      getUP(i,q[tail-1],q[tail-2])*getDOWN(j,q[tail-1])) tail--;
                q[tail++] = j;
            }
        }
        printf("Case %d: %d\n", kase, dp[m][n]);
    }
}