【DP+四边不等式优化】Division

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

23 21 2 44 24 7 10 1

 

Sample Output

Case 1: 1Case 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 5010int 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;}