【POJ 3977 Subset】 二进制枚举+贪心

POJ 3977

• 给定n个数，求一个子集(非空)
• 使得子集内元素和的绝对值最小
• n ≤ 35

首先看到35 很容易想到二进制枚举
那么假如左边取了 x 那么右边肯定去找第一个大于-x的
然后情况分成 1.左边
2.右边
3.左边加右边
注意题目还说如果值相同 要元素个数最小 一开始没看到调了半天

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);

/*namespace sgt
{
    #define mid ((l+r)>>1)

    #undef mid
}*/
long long a[45];
map<ll ,int > mp;
long long Abs(long long x)
{
    return x < 0?-x:x;
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n;
    while(scanf("%d",&n)&&n)
    {
        mp.clear();
        for(int i = 1;i<=n;++i)
        {
            scanf("%lld",&a[i]);
        }
        int tmp = n/2;
        pair<long long ,int > minn = make_pair(Abs(a[1]),1);
        for(int i = 1;i<(1<<tmp);++i)
        {
            int cnt = 0;
            ll now = 0;
            for(int j = 0;j<tmp;++j)
            {
                if(i&(1<<j))
                {
                    cnt++;
                    now+=a[j+1];
                }
            }
            if(minn.first>Abs(now)) minn = make_pair(Abs(now),cnt);
            else if(minn.first==Abs(now)) minn = make_pair(Abs(now),min(cnt,minn.second));
            if(mp.find(now)!=mp.end()) mp[now] = min(cnt,mp[now]);
            else mp[now] = cnt;
        }
        for(int i = 1;i<(1<<(n-tmp));++i)
        {
            int cnt = 0;
            ll now = 0;
            for(int j = 0;j<(n-tmp);++j)
            {
                if(i&(1<<j))
                {
                    cnt++;
                    now += a[j+1+tmp];
                }
            }
            if(minn.first>Abs(now)) minn = make_pair(Abs(now),cnt);
            else if(minn.first==Abs(now)) minn = make_pair(Abs(now),min(cnt,minn.second));
            map<ll,int > ::iterator it;
            it = mp.lower_bound(-now);
            if(it!=mp.end())
            {
                if(minn.first>Abs(now+it->first)) minn = make_pair(Abs(now+it->first),cnt+it->second);
                else if(minn.first==Abs(now+it->first)) minn = make_pair(Abs(now+it->first),min(cnt+it->second,minn.second));
            }
            if(it!=mp.begin())
            {
                it--;
                if(minn.first>Abs(now+it->first)) minn = make_pair(Abs(now+it->first),cnt+it->second);
                else if(minn.first==Abs(now+it->first)) minn = make_pair(Abs(now+it->first),min(cnt+it->second,minn.second));
            }
        }
        printf("%lld %d\n",minn.first,minn.second);
    }
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}