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小白dp uva 10271 - Chopsticks

編輯：C++入門知識

Problem C
Chopsticks
Input: Standard Input
Output: Standard Output

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.

Input

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 L_i on the next line in non-decreasing order indicating the lengths of the chopsticks.(1<=L_i<=32000).

Output

For each test case in the input, print a line containing the minimal total badness of all the sets.

Sample Input

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

Sample Output

Note

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

______________________________________________________________

題意：

從給定的組合中找出 (A,B,C)這樣的k+8個組合，每個組合滿足(A<=B<=C)，badness為 (A-B)^2，使得總badness最小。

思路：

開始覺得C不用考慮，那就很好轉移了，發現沒有這麼簡單，因為C要滿足最大，直接轉移會使在你把A、B選好後，沒有那麼多C滿足你的條件。從小到大轉移不好控制，那就從大到小排序後轉移了。

dp[i][j]-前j個數能選出i個組合 i-選第幾個組合 j-前j個數

那麼有

dp[i][j]=min(dp[i][j-2],dp[i-1][j-2]+(a[j]-a[j-1])^2)；

當然取第二個時需滿足 j>=3*i 啦。

代碼：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 5005
#define MAXN 100005
#define mod 1000000007
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 0.000001
typedef long long ll;
using namespace std;

int n,m,ans,cnt,tot,flag;
int a[maxn],dp[2][maxn];

void solve()
{
    int i,j,t,p=0;
    memset(dp,0x3f,sizeof(dp));
    for(i=0;i<=n;i++)
    {
        dp[0][i]=0;
    }
    for(i=1;i<=m;i++)
    {
        for(j=0;j<3*i;j++) dp[p^1][j]=INF;
        for(j=3*i;j<=n;j++)
        {
              dp[p^1][j]=dp[p^1][j-1];
              if(j>=3*i)
              {
                  dp[p^1][j]=min(dp[p^1][j],dp[p][j-2]+(a[j]-a[j-1])*(a[j]-a[j-1]));
              }
        }
        p=p^1;
    }
    ans=INF;
    for(j=3*m;j<=n;j++)
    {
        ans=min(ans,dp[p][j]);
    }
}
bool cmp(int x,int y)
{
    return x>y;
}
int main()
{
    int i,j,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&m,&n);
        m+=8;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        sort(a+1,a+n+1,cmp);
        solve();
        printf("%d\n",ans);
    }
    return 0;
}