TOJ 2865 Chopsticks DP

2865: Chopsticks

묘사
In China, people use a pair of chopsticks to get food on the table, but Mr. L is a bit different. He uses a set of three chopsticks -- one pair, plus an EXTRA long chopstick to get some big food by piercing it through the food. As you may guess, the length of the two shorter chopsticks should be as close as possible, but the length of the extra one is not important, as long as it's the longest. To make things clearer, for the set of chopsticks with lengths A,B,C(A<=B<=C), (A-B)^2 is called the 'badness' of the set.
It's December 2nd, Mr.L's birthday! He invited K people to join his birthday party, and would like to introduce his way of using chopsticks. So, he should prepare K+8 sets of chopsticks(for himself, his wife, his little son, little daughter, his mother, father, mother-in-law, father-in-law, and K other guests). But Mr.L suddenly discovered that his chopsticks are of quite different lengths! He should find a way of composing the K+8 sets, so that the total badness of all the sets is minimized.
입력
The first line in the input contains a single integer T, indicating the number of test cases(1<=T<=20). Each test case begins with two integers K, N(0<=K<=1000, 3K+24<=N<=5000), the number of guests and the number of chopsticks. There are N positive integers Li on the next line in non-decreasing order indicating the lengths of the chopsticks.(1<=Li<=32000).
출력
For each test case in the input, print a line containing the minimal total badness of all the sets.
샘플 입력
1 1 40 1 8 10 16 19 22 27 33 36 40 47 52 56 61 63 71 72 75 81 81 84 88 96 98 103 110 113 118 124 128 129 134 134 139 148 157 157 160 162 164
샘플 출력
23
프롬프트
For the sample input, a possible collection of the 9 sets is: 8,10,16; 19,22,27; 61,63,75; 71,72,88; 81,81,84; 96,98,103; 128,129,148; 134,134,139; 157,157,160
제목: L씨는 n개의 젓가락에서 k+8쌍을 골라야 한다. 그의 젓가락은 한 쌍에 3개의 젓가락 abc로 구성되어 있는데 그 중에서 c가 가장 길고 (a-b)*(a-b)가 이 젓가락의badness값을 구성하여 k+8쌍의 젓가락을 구성하는 최소badness값을 구한다.
사고방식: c가 가장 크고 ab가badness값을 구성하기 때문에 전체 수조를 큰 수조에서 작은 수조로 배열할 수 있다. 수조 dp[i][j]는 i개의 젓가락 중에서 j쌍을 선택한다.
전이 방정식: f[i][j]=min(f[i-1][j], f[i-2][j-1]+(a[i]-a[i-1])*(a[i]-a[i-1]))))*(a[i]-a[i-1])))).

#include 
#include 
#include 
using namespace std;
int a[5500],f[5500][1500];
//f[i][j]   i    j 
bool cmp(int x,int y)
{
	return x>y;
}
int main()
{
	int t,n,k,i,j;
    scanf("%d",&t);
    while(t--)
    {
        memset(a,0,sizeof(a));
        scanf("%d%d",&k,&n);
        for(i=1;i<=n;i++)
        scanf("%d",&a[i]);
        sort(a+1,a+n+1,cmp);
        for(i=1;i<=n;i++)
        {
            f[i][0]=0;
            for(j=1;j<=k+8;j++)
            f[i][j]=10000000;
        }
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=k+8;j++)
            {
                if(3*j<=i)
                f[i][j]=min(f[i-1][j],f[i-2][j-1]+(a[i]-a[i-1])*(a[i]-a[i-1]));
            }
        }
        printf("%d
",f[n][k+8]);
    }
}