hdu 1500 Chopsticks (dp)

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我是靠谱客的博主美好火，最近开发中收集的这篇文章主要介绍hdu 1500 Chopsticks (dp)，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

Description

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

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

另dp[i][j]表示前i个人在前j支筷子里面选；

dp[i][j]=min(dp[i][j-1],dp[i-1][j-2]+(ma[i]-ma[i-1])*(ma[i]-ma[i-1]))。

因为首先要保证有解，所以先dp[i][i*3]=dp[i-1][i*3-2]+(ma[i][i*3-1]-ma[i][i*3])*(ma[i][i*3-1]-ma[i][i*3]);

#include <iostream>
#include <cstring>
using namespace std;

int dp[1100][5001];

int ma[5001];

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        memset(dp,0,sizeof(dp));
        int k,n;
        cin>>k>>n;
        k+=8;
        for(int i=n;i>=1;i--)
            cin>>ma[i];
        for(int i=1;i<=k;i++)
        {
            dp[i][i*3]=dp[i-1][i*3-2]+(ma[i*3-1]-ma[i*3])*(ma[i*3-1]-ma[i*3]);
            for(int j=i*3+1;j<=n;j++)
                dp[i][j]=min(dp[i][j-1],dp[i-1][j-2]+(ma[j-1]-ma[j])*(ma[j-1]-ma[j]));
        }
        cout<<dp[k][n]<<endl;
    }
}