POJ 3977 Subset

最新推荐文章于 2025-03-19 20:53:34 发布

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最新推荐文章于 2025-03-19 20:53:34 发布

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原文链接：http://www.cnblogs.com/DOlaBMOon/p/7345348.html

本文介绍了一种解决特定子集和问题的方法，该问题要求找到一个非空子集，使得其元素之和的绝对值最小，并且在多个解中选择元素最少的那个。采用折半搜索策略，将列表分为两部分，分别计算所有可能的子集和，通过排序和二分查找优化解决方案。

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Subset

Time Limit: 30000MS		Memory Limit: 65536K
Total Submissions: 4862		Accepted: 892

Description

Given a list of N integers with absolute values no larger than 10 ¹⁵, find a non empty subset of these numbers which minimizes the absolute value of the sum of its elements. In case there are multiple subsets, choose the one with fewer elements.

Input

The input contains multiple data sets, the first line of each data set contains N <= 35, the number of elements, the next line contains N numbers no larger than 10 ¹⁵ in absolute value and separated by a single space. The input is terminated with N = 0

Output

For each data set in the input print two integers, the minimum absolute sum and the number of elements in the optimal subset.

Sample Input

20 100 -100

Sample Output

10 1

0 2

Solution:

折半搜索,处理出前半组数所能组成的所有和,以及后半组数所能组成的所有和,先用这些和更新答案,然后再把这两组和分别排序,对于第一组中的每一个数,在另一组二分查找到能和他组成的绝对值最小的和,再更新答案.

Code:

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
using namespace std;
#define pii pair<ll,int>
#define ll long long
const int MAXN=35;
const int MAXTOT=300000;
int n;
ll a[MAXN+10];
pii res1[MAXTOT+10],res2[MAXTOT+10];
int tot1,tot2;
int cnt;ll sum;
ll myabs(ll a){return (a<0)?(-a):a;}
void solve(int k,int n,pii* res,int& tot)
{
	if(k>n){if(cnt)res[++tot]=make_pair(sum,cnt);return;}
	solve(k+1,n,res,tot);
	++cnt;sum+=a[k];
	solve(k+1,n,res,tot);
	--cnt;sum-=a[k];
}
pii tmp[MAXTOT+10];
void doit(pii* res,int& tot)
{
	int now=1;
	for(int i=2;i<=tot;++i)
	{
		if(res[i].first!=res[i-1].first)tmp[++now]=res[i];
	}
	tot=now;
	for(int i=2;i<=tot;++i)res[i]=tmp[i];
}
void cmp(ll& ans,int& cnt,ll res,int t)
{
	res=myabs(res);
	if(ans==res)cnt=min(cnt,t);
	else if(ans>res)
	{
		ans=res;
		cnt=t;
	}
}
int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);
	while(scanf("%d",&n),n)
	{
		for(int i=1;i<=n;++i)scanf("%lld",a+i);
		if(n==1){printf("%lld 1\n",myabs(a[1]));continue;}
		int mid=(n>>1);
		cnt=sum=tot1=tot2=0;
		solve(1,mid,res1,tot1);
		solve(mid+1,n,res2,tot2);
		sort(res1+1,res1+tot1+1);
		sort(res2+1,res2+tot2+1);
		doit(res1,tot1);
		doit(res2,tot2);
		ll ans=1e18;int cnt;
		for(int i=1;i<=tot1;++i)
		{
			int left=1,mid,right=tot2,res=1;
			while(left<=right)
			{
				mid=(left+right)>>1;
				if(res2[mid].first<=-res1[i].first)
				{
					res=mid;
					left=mid+1;
				}
				else right=mid-1;
			}
			cmp(ans,cnt,res2[res].first+res1[i].first,res2[res].second+res1[i].second);
			if(res<tot2)cmp(ans,cnt,res2[res+1].first+res1[i].first,res2[res+1].second+res1[i].second);
		}
		for(int i=1;i<=tot1;++i)cmp(ans,cnt,res1[i].first,res1[i].second);
		for(int i=1;i<=tot2;++i)cmp(ans,cnt,res2[i].first,res2[i].second);
		printf("%lld %d\n",myabs(ans),cnt);
	}
	return 0;
}

转载于:https://www.cnblogs.com/DOlaBMOon/p/7345348.html