[动态规划]Zju1234--Chopsticks

Chopsticks

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Time Limit: 10 Seconds      Memory Limit: 32768 KB

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

23

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

题目大意：有一个人发明了一种新的用筷子的方法，每次用三根筷子，筷子长度不一（A<=B<=C），规定三根筷子的“不合适度”用(A-B)^2来表示。现在有n根长度不一的筷子，问选出(m+8)副筷子的最小不合适度之和是多少。输入时按照筷子长度非降序输入。

 

分析：经典动态规划问题。不难得出，若第i根筷子是一组筷子中最短的，最优方案中该组筷子里第二短的一定是第i+1根筷子。于是能得出以下的动态规划：

定义f[i,j]表示后i根筷子组成j组筷子时的最小不合适度，显然初值为f[i,0]=0,f[n-2,1]=(a[n-2]-a[n-1])^2。

f[i,j]=min(f[i+1,j],f[i+2,j-1]+(a[i]-a[i+1])^2)。注意n-i+1要大于等于3*j。还有就是要倒推。

 

codes：

var t,n,m:longint; f:array[0..5000,0..1008] of longint; a:array[0..5000] of longint;procedure init; var i:longint; begin readln(n,m); n:=n+8; for i:=1 to m do read(a[i]); end;function g(x:longint):int64; begin exit(sqr(a[x]-a[x+1])); end;function min(a,b:int64):int64; begin if a>b then a:=b; min:=a; end;procedure dp; var i,j:longint; begin fillchar(f,sizeof(f),127); for i:=1 to m do f[i,0]:=0; f[m-2,1]:=g(m-2); for i:=m-3 downto 1 do for j:=1 to n do if j*3>(m-i+1) then break else f[i,j]:=min(f[i+1,j],f[i+2,j-1]+g(i)); writeln(f[1,n]); end;begin readln(t); while t>0 do begin dec(t); init; dp; end;end.