POJ 3977

2017-08-01 21:45:19

writer：pprp

题目：

• 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

---

AC代码如下：（难点：枚举出sum）

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>


using namespace std;

const int maxn = 40;
const int INF = 0x3f3f3f3f;
typedef long long ll;
int n;
ll a[maxn];
int Min = INF;


ll ll_abs(ll x)
{
    return x>=0?x:-x;
}

int main()
{
    while(cin >> n && n)
    {
        memset(a,0,sizeof(a));
        for(int i = 0 ; i < n ; i++)
        {
            cin >> a[i];
        }
        map<ll, int>mp; //sum -> cnt
        pair<ll,int>ans(ll_abs(a[0]),1);   //储存最优解

        for(int i = 0 ; i < (1<<(n/2)) ; i++)
        {

            ll sum = 0;
            int cnt = 0;
            for(int j = 0 ; j < (n/2) ; j++)
            {
                if((i>>j)&1)
                {
                    sum += a[j];
                    cnt++;
                }
            }

            if(cnt == 0)
                continue;
            ans = min(ans,make_pair(ll_abs(sum),cnt));

            map<ll, int>::iterator it = mp.find(sum);

            if(it != mp.end())
            {
                //取更小的元素个数
                it->second = min(it -> second, cnt);
            }
            else
                mp[sum] = cnt;

        }

        for(int i = 0 ; i < (1 << (n - n / 2)) ; i++)
        {
            ll sum = 0;
            int cnt = 0;
            for(int j = 0 ; j < (n - n / 2) ; j++)
            {
                if((i>>j)&1)
                {
                    sum += a[n / 2 + j];
                    cnt++;
                }
            }

            if(cnt == 0)  continue;
            ans = min(ans, make_pair(ll_abs(sum), cnt));
            
            map<ll, int> ::iterator it = mp.lower_bound(-sum);

            if(it != mp.end())
                ans = min(ans, make_pair(ll_abs(it->first + sum), it->second + cnt));

            if(it != mp.begin())
            {
                  it--;
                  ans = min(ans, make_pair(ll_abs(it->first + sum), it->second + cnt) );
            }
        }
        cout << ans.first <<" " << ans.second << endl;
    }

    return 0;
}