【DP+四边不等式优化】Division

最新推荐文章于 2024-05-10 20:52:40 发布

ACM_Ted

最新推荐文章于 2024-05-10 20:52:40 发布

阅读量592

点赞数

CC 4.0 BY-SA版权

分类专栏： ACM 文章标签：优化 integer input output fun each

本文链接：https://blog.youkuaiyun.com/ACM_Ted/article/details/7492411

ACM 专栏收录该内容

149 篇文章

订阅专栏

本文解析了HDU 3480分割问题，通过使用动态规划结合四边不等式优化的方法求解最小总成本。输入包含多个测试案例，输出为按最优方式分割集合后的最小总成本。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Division

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

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

Sample Output

  
   Case 1: 1
Case 2: 18

DP的四边不等式优化，看了一个晚上，终于对理论上的有了一点理解。

这题还是参考别人的代码改的。

#include<cstdio>
#include<algorithm>
using namespace std;
#define N 10010
#define M 5010
int val[N];
int dp[M][N];//dp[i][j]代表i堆,前j个数的最优解 
int s[M][N];//s[i,j]为dp[i,j]的决策量，即dp[i,j]=dp[i,s[i,j]]+(val[j]-val[s[i,j]+1])^2)
//dp[i][j]=min(dp[i-1][k]+(val[j]-val[k+1])^2)
int n,m;
int fun(int x)
{
    return x*x;
}
int main()
{
    int t,ti=1;
    scanf("%d",&t);
    for(;t--;)
    {
        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)
        {
            dp[1][i]=fun(val[i]-val[1]);//前i个数分为1堆的最优解 
            s[1][i]=0;//dp[1][i]的最优决策是dp[i][0]+(val[i]-val[1])^2 
            dp[i][i]=0;//前i个数分为i堆的最优解 
        }
        for(int i=2;i<=m;++i)
        {
            s[i][n+1]=n-1;
            for(int j=n;j>i;--j)
            {
                dp[i][j]=-1;
                for(int k=s[i-1][j];k<=s[i][j+1]&&k<j;++k)//四边不等式优化 
                {
                    int temp=dp[i-1][k]+fun(val[j]-val[k+1]);
                    if(dp[i][j]==-1||dp[i][j]>temp)//保留最优解 
                    {
                        dp[i][j]=temp;
                        s[i][j]=k;
                    }
                }
            }
        }
        printf("Case %d: %d\n",ti++,dp[m][n]);
    }
    return 0;
}