uva10271 - Chopsticks

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 thelength 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 tointroduce 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, andK 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

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

  有K+8个人，一个人3个筷子？=。=好奇怪。。要求这三个筷子A,B,C，A<=B<=C，并且有一个值(A-B)^2，最后要使所有人的这个值加起来最小，给出的数据是升序。

  这个题看起来很复杂的样子，条件很多。其中有很重要的一点，AB一定是相邻的！（可以数学证明一下。。相邻的情况是最优的）要发现了这一点才能往后做。设dp[i][j]是前j个数组成i组的最好情况，dp[i][j]=min(dp[i][j-1],dp[i-1][j-2]+(a[j]-a[j-1])*(a[j]-a[j-1]))，这个很巧妙，要保证的是j至少有3*i个，只要按降序排列，就排除了C的影响，因为如果前面的都比a[i],a[i-1]大，任取一个作为C都可以，但如果是升序就不行，因为如果把i,i-1作为A,B了就没有更大的C了。

#include<cstring>#include<cstdio>#include<iostream>#include<cmath>#include<algorithm>#include<queue>#define INF 0x3f3f3f3fusing namespace std;int a[5010];int dp[1010][5010];int main(){   freopen("in.txt","r",stdin);    int T,N,K;    scanf("%d",&T);    while(T--){        scanf("%d%d",&K,&N);        K+=8;        int i,j;        memset(dp,INF,sizeof(dp));        for(i=N;i>0;i--){            scanf("%d",&a[i]);            dp[0][i]=0;        }        for(i=1;i<=K;i++)            for(j=3*i;j<=N;j++)            dp[i][j]=min(dp[i][j-1],dp[i-1][j-2]+(a[j]-a[j-1])*(a[j]-a[j-1]));        printf("%d\n",dp[K][N]);    }    return 0;}