2020牛客暑期多校训练营（第三场）E.Two Matchings

2020牛客暑期多校训练营（第三场）E.Two Matchings

题目链接

题目描述

A permutation of length n is an array $p=[p_1,p_2,\dots ,p_n]$, which contains every integer from 1 to n (inclusive) and each number appears exactly once. For example, $p=[3,1,4,6,2,5]$ is a permutation of length 6.

Let’s call a permutation is a matching if and only if $p_i\ne i$ and $p_{p_i}=i$
for all valid i.

You are given an array $a=[a_1,a_2,\dots ,a_n](0\le a_i\le 10^9,n\ge 4$ and n is even). Define the cost of a permutation is $(\sum _{i=1}^{n}abs(a_i-a_{p_i}))/2$.

Define two matchings p, q are combinable if and only if $p_i\ne q_i$ for all i from 1 to n.

Please find two combinable matchings such that the sum of the cost of these two matchings is as small as possible. Output the sum.

输入描述:

The first line contains one integer t $(1\le t\le 5\times 10^4)$ — the number of test cases.

Each test contains two lines. The first line contains one integer $n(n\ge 4n\ge 4$ and n is even), The second line contains n integers $a_1,a_2,\dots ,a_n(0\le a_i\le 10^9)$.

The sum of n across the test cases doesn’t exceed $2\times 10^5$.

输出描述:

For each test, output one line which contains only one integer representing the answer of this test.

示例1

输入

2
4
0 8 0 0
6
3 1 4 1 5 9

输出

16
16

比赛时打死也想不到是 DP，果然，想不到方法的题目通通是 DP
网上大部分代码都是直接加的，这边提供一个减的思路，也是题解做法：

• 此题等价于找一些长度为偶数的环使得这些环恰通过每个点一次， 且所有边的总权重最少。
• 2k 个点所构成的环的权重和的最小值为最大的权重减最小的权重。
• 先把所有点的按照权重排序，最佳解一定是出现在每个环都是由排序后连续的点构成。
• 若某个环长度≥ 8，总是可以拆成一些长度为 4 或 6 的环且总边权更小。
  于是我们就可以把点的权重排序后， 对于前 2k 个点当作子题来动态规划啦，转移时只要枚举最后一个点所在的环长度是 4 还是 6 即可。

AC代码如下：

#include
using namespace std;
const int N=2e5+5;
int t,n,x,a[N],dp[N];
int main()
{
	scanf("%d",&t);
	while(t--){
        scanf("%d",&n);
        for(int i=1;i<=n;i++) scanf("%d",&a[i]),dp[i]=0;
        sort(a+1,a+1+n);
        for(int i=5;i<=n-3;i+=2) dp[i]=max(dp[i-2],dp[i-4]+a[i]-a[i-1]);
        printf("%d\n",2*(a[n]-a[1])-2*dp[n-3]);
	}
}