Chopsticks - UVa 10271 dp

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本文介绍了一个有趣的算法问题——如何从不同长度的筷子中选择最佳组合，以使每组三只筷子（两短一长）的长度差平方和最小。通过动态规划的方法解决了这一问题，并给出了详细的实现代码。

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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

题意：从n只筷子里取出k+8组筷子，每组有三只筷子，使得每组的两只短的差的平方和最小。

思路：首先取的两根短的筷子一定是相邻的，然后排序按从大到小排，dp[i][j]表示到第i只筷子取出j组的最小值。

dp[i][j]=min(dp[i-1][j],dp[i-2][j-1]+(num[i]-num[i-1])*(num[i]-num[i-1]));

当i>.=3*j时，那么只考虑两只相邻的短筷子一定是有解的。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[5010][1010],num[5010];
int main()
{ int t,n,m,i,j,k;
  scanf("%d",&t);
  while(t--)
  { scanf("%d%d",&m,&n);
    m+=8;
    memset(dp,0,sizeof(dp));
    for(i=1;i<=n;i++)
     for(j=1;j<=m;j++)
      dp[i][j]=1000000000;
    for(i=n;i>=1;i--)
     scanf("%d",&num[i]);
    for(i=3;i<=n;i++)
     for(j=1;j<=m;j++)
      if(i>=j*3)
       dp[i][j]=min(dp[i-1][j],dp[i-2][j-1]+(num[i]-num[i-1])*(num[i]-num[i-1]));
    printf("%d\n",dp[n][m]);
}
}