HDU3480 Division DP四边形不等式优化

Division

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 999999/400000 K (Java/Others)
Total Submission(s): 5868    Accepted Submission(s): 2332

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

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题意：
将n个数，分成m个子集，使得每个子集的最大值与最小值的平方差之和最小

分析：

DP四边形不等式优化

dp[i][j]=dp[i-1][k-1]+w[k][j]，由于w[i+1][j]-w[i][j]=(a[i]-a[i+1])*(2*a[j]-a[i+1]-a[i])，这个表达式是随着j的增加单调递减的，所以就有w[i+1][j+1]-w[i][j+1]<=w[i+1][j]-w[i][j]，也就是w[i][j]+w[i+1][j+1]<=w[i][j+1]+w[i+1][j]，满足四边形不等式优化的条件，我们就可以放心地使用四边形不等式优化dp了。

对于dp[i][j]的决策s[i][j]，有s[i][j-1]<=s[i][j]<=s[i+1][j]，在dp[i][j]求出来之前要先求出dp[i][j-1]、dp[i+1][j]，可以按照长度递增来递推，具体见代码，还有注意初始化。

#include
using namespace std;
const int maxn=1e4+10;
#define inf 0x3f3f3f3f
int dp[maxn>>1][maxn];//dp[i][j]表示在【i，j】区间上的最优值
int s[maxn>>1][maxn];//s[i][j]表示dp[i][j]取最优值时的下标
int a[maxn];//n个数字
int n,m;//n个数，分成m个集合
int ans;//答案
void solve()
{
    memset(dp,inf,sizeof(dp));
    for(int i=0;i<=m;i++)
    {
        dp[i][i]=0;
        s[i][i]=i;
    }
    for(int i=m+1;i<=n;i++)
        s[m+1][i]=i;
    for(int len=1;lenn) break;//越界
            for(int k=s[i][j-1];k<=s[i+1][j];k++)//枚举区间断点
            {
                int temp=dp[i-1][k-1]+(a[k]-a[j])*(a[k]-a[j]);
                if(dp[i][j]>temp)//更新操作
                    dp[i][j]=temp,s[i][j]=k;
            }
        }
    }
    ans=dp[m][n];//n个数，分成m个集合的最优值
}
int main()
{
    int T,Case=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        sort(a+1,a+n+1);
        solve();
        printf("Case %d: %d\n",Case++,ans);
    }
    return 0;
}