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

题目描述

输入描述

The first line contains one integer t (1≤t≤5×10^4) --- the number of test cases.

Each test contains two lines. The first line contains one integer n (n≥4n \ge 4n≥4 and n is even), The second line contains n integers a1​,a2​,…,an​ (0≤ai​≤10^9).

The sum of n across the test cases doesn't exceed 2×10^5.

输出描述

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

输入

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

输出

16
16

说明

In the first test, one possible combinable matchings with the minimum sum of cost are [2,1,4,3] and [4,3,2,1]. The cost of the first matching is (|0-8|+|8-0|+|0-0|+|0-0|)/2=8, the cost of the second matching is (|0-0|+|8-0|+|0-8|+|0-0|)/2=8.So their sum is 8+8=16.

题目大意

题解

我们可以通过排序，之后更方便匹配，同时以++--（长度为4时）++-+--（长度为6时）最优，之后可以使用dp求值。

AC代码

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include<bits/stdc++.h>
#define INF 1ll<<60
using namespace std;
long long a[202020],dp[202020];
int main()
{
	long long ans,T,n;
	scanf("%lld",&T);
	while(T--)
	{
		scanf("%lld",&n);
		ans=0;
		for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
		sort(a+1,a+1+n);
		for(int i=2;i<=n;i+=2) ans+=a[i]-a[i-1];
		dp[2]=INF;
		dp[4]=a[4]+a[3]-a[2]-a[1];
		dp[6]=a[6]+a[5]-a[4]+a[3]-a[2]-a[1];
		for(int i=8;i<=n;i+=2)
			dp[i]=min(dp[i-4]+a[i]+a[i-1]-a[i-2]-a[i-3],dp[i-6]+a[i]+a[i-1]-a[i-2]+a[i-3]-a[i-4]-a[i-5]);
		printf("%lld\n",ans+dp[n]);
	}
}