UVA 题目10271 - Chopsticks（DP）

最新推荐文章于 2016-03-07 21:55:00 发布

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最新推荐文章于 2016-03-07 21:55:00 发布

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文章详细介绍了在中国使用筷子的传统方式，并引入了一种创新的三根筷子使用方法，通过增加一根较长的筷子来方便获取较大食物。文章还探讨了如何根据筷子的长度差异来最小化食物获取的复杂度，提供了实际应用案例和计算方法。

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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.
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
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

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
#define INF 0x3f3f3f3f3f3f3f
int dp[5010][1010];
int val[5010];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int m,n;
        scanf("%d%d",&m,&n);
        m+=8;
        int i,j;
        memset(dp,INF,sizeof(dp));
        for(i=n;i>=1;i--)
        {
            dp[i][0]=0;
            scanf("%d",&val[i]);
        }
        for(i=3;i<=n;i++)
            for(j=1;j<=m;j++)
            {
                if(i>=j*3&&dp[i-2][j-1]!=INF)
                    dp[i][j]=min(dp[i-1][j],dp[i-2][j-1]+(val[i-1]-val[i])*(val[i-1]-val[i]));
            }
        printf("%d\n",dp[n][m]);
    }
}