Codeforces GYM 100340A Cookies_codeforces gym cookies-优快云博客

本文链接：https://blog.youkuaiyun.com/Rlt1296/article/details/68926833

本文介绍了一个关于如何公平分配有限数量饼干的问题。圣诞老人需要将饼干分给孩子们，每个孩子有不同的贪婪度，目标是最小化孩子们的总体不高兴值。通过贪心算法和动态规划求解最优解。

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Santa Claus is planning to bring gifts to n children. He has m cookies and is planning to divide them to n piles. However, as usually problems come unexpected. The child gets unhappy if somebody gets more cookies than him. Each child is characterized by his greediness, the greediness of the i-th child is gi. The unhappines of the i-th child is equal to giai where ai is the number of children that get more cookies than him. Now Santa wants to divide cookies in such a way that the total unhappiness is minimized. Each child must get at least one cookie. Santa would like to give away all m cookies he has. Help him to do so. (1 ≤ n ≤ 30, n ≤ m ≤ 5000)

首先，一定是贪心，更贪婪的孩子，得到的糖果多。

之后，设f[i][j]表示前i个小孩分配j个糖果的最小不高兴值。

则1如果这个小孩分了>1颗糖，所有小孩都减一，不高兴值不变。f[i][j]=f[i][j-i];

2如果这个小孩分了一颗糖，枚举分得一颗糖的人数k.

f[i][j]=f[i-k][j-k]+(i-k)*(s[i]-s[i-k]);

之后方案数，记录前驱，模拟即可。

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int n,m,g[35],a[35],s[35],f[35][5005],pre[35][5005],ans[35];
bool b[35][5005];
//f[i][j]前i个小孩分配j个糖果的最小不高兴值 
bool cmp(int c,int d)
{
	return g[c]>g[d];
}
int main()
{
	freopen("cookies.in","r",stdin);
    freopen("cookies.out","w",stdout);
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&g[i]);
		a[i]=i;
	}
	sort(a+1,a+n+1,cmp);
	for(int i=1;i<=n;i++)
		s[i]=s[i-1]+g[a[i]];
	memset(f,0x3f,sizeof(f));
	f[0][0]=0;
	for(int i=1;i<=n;i++)
		for(int j=i;j<=m;j++)
		{
			f[i][j]=f[i][j-i];
			for(int k=1;k<=i;k++)//从后向前共有k个小孩获得了1个糖果 
				if(f[i][j]>f[i-k][j-k]+(i-k)*(s[i]-s[i-k]))
				{
					f[i][j]=f[i-k][j-k]+(i-k)*(s[i]-s[i-k]);
					b[i][j]=1;
					pre[i][j]=k;
				}
		}
	printf("%d\n",f[n][m]);
	int p=n,t=m;
	while(p)
	{
		if(b[p][t])
		{
			int x=pre[p][t];
			for(int i=p-x+1;i<=p;i++)
				ans[a[i]]++;
			t-=x;
			p-=x;
		}
		else
		{
			for(int i=1;i<=p;i++)
				ans[a[i]]++;
			t-=p;
		}
	}
	for(int i=1;i<=n-1;i++)
		printf("%d ",ans[i]);
	printf("%d\n",ans[n]);
	return 0;
}