POJ 3977 Subset(二分+折半枚举）

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解决一个给定整数列表的子集求和问题，寻找绝对值最小的非空子集和，通过两分搜索和排序优化算法实现。

Subset

Time Limit: 30000MS		Memory Limit: 65536K
Total Submissions:5715		Accepted: 1081

Description

Given a list of N integers with absolute values no larger than 10¹⁵, 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 10¹⁵ 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

1
10
3
20 100 -100
0

Sample Output

10 1
0 2

#include<cstdio>
#include<algorithm>
#include<map>
using namespace std;
typedef long long ll;
ll a1[22],a2[22],d[2000003];
map<ll,int>mp;//去重
struct node
{
    int num;
    ll cost;
} c[2000003];
bool cmp(const node& a,const node& b)
{
    return a.cost<b.cost;
}
ll abs1(ll x)
{
    if(x<0)x=-x;
    return x;
}
int main()
{
    int n;
    while(~scanf("%d",&n),n)
    {
        int i,j,k1=n/2,k2=n-k1,y=0;
        for(i=0; i<k1; i++)
            scanf("%lld",&a1[i]);
        for(i=0; i<k2; i++)
            scanf("%lld",&a2[i]);
        for(int t=0; t<1<<k1; t++)
        {
            ll cost=0;int num=0;
            for(i=0; i<k1; i++)
            {
                if(t>>i&1)
                {
                    cost+=a1[i];
                    num++;
                }
            }
            if(mp[cost])c[mp[cost]].num=min(c[mp[cost]].num,num);
            else
            {
                mp[cost]=y;
                c[y].cost=cost;
                c[y++].num=num;
            }
        }
        sort(c,c+y,cmp);
        y--;
        for(i=0; i<=y; i++)
            d[i]=c[i].cost;
        ll mi=1e17;
        int sum=0;
        for(int t=0; t<1<<k2; t++)
        {
            ll cost=0;
            int num=0;
            for(i=0; i<k2; i++)
            {
                if(t>>i&1)
                {
                    cost+=a2[i];
                    num++;
                }
            }
            int id;
            ll ta;
            id=lower_bound(d+1,d+y+1,-cost)-d;
            if(id<=y)//防止越界
            {
                ta=abs1(cost+d[id]);
                num=num+c[id].num;
                if(ta<mi&&num)mi=ta,sum=num;
                else if(ta==mi&&num)sum=min(sum,num);
                num-=c[id].num;
            }
            if(id-1>=0)//判断两个边界情况
            {
                id--;
                ta=abs1(cost+d[id]);
                num+=c[id].num;
                if(ta<mi&&num)mi=ta,sum=num;
                else if(ta==mi&&num)sum=min(sum,num);
            }
        }
        printf("%lld %d\n",mi,sum);
        mp.clear();
    }
    return 0;
}