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最新推荐文章于 2021-05-26 21:03:23 发布

原创最新推荐文章于 2021-05-26 21:03:23 发布 · 240 阅读

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状态Dp[i][j]为前i件物品选j对的最优解
当i=j*2时,只有一种选择即 Dp[i-2][j-1]+(w[i]-w[i-1])^2

当i>j*2时,Dp[i][j] = min(Dp[i-1][j],Dp[i-2][j-1]+(w[i]-w[i-1])^2)

以为要开long long，结果爆了内存；

没想到排序，wa了好几次＝＝

//  Created by Chenhongwei in 2015.
//  Copyright (c) 2015 Chenhongwei. All rights reserved.

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <climits>
#include <queue>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <vector>
#include <sstream>
#include <algorithm>
using namespace std;
const int inf=1e9;
const int maxn=1e5+100;
typedef long long ll;
typedef unsigned long long ull;
int w[2020];
int dp[2004][1002];
int main()
{	
	//ios::sync_with_stdio(false);
	// freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);
	int n,k;
	while(scanf("%d%d",&n,&k)!=EOF)
	{
		memset(dp,0x3f,sizeof dp);
		for(int i=1;i<=n;i++)
			scanf("%d",&w[i]);
		sort(w+1,w+n+1);
		for(int i=0;i<=n;i++)
			dp[i][0]=0;
		for(int i=1;i<=n;i++)
			for(int j=1;j<=k;j++)
			{
				if(j*2>i)
					continue;
				else
					dp[i][j]=min(dp[i-1][j],dp[i-2][j-1]+(w[i]-w[i-1])*(w[i]-w[i-1]));
			}
		printf("%d\n",dp[n][k]);
	}
	return 0;
}