POJ - 3977 Subset（二分+折半枚举）

题意：有一个N（N <= 35）个数的集合，每个数的绝对值小于等于10¹⁵，找一个非空子集，使该子集中所有元素的和的绝对值最小，若有多个，则输出个数最小的那个。

分析：

1、将集合中的元素分成两半，分别二进制枚举子集并记录子集所对应的和以及元素个数。

2、枚举其中一半，二分查找另一半，不断取最小值。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-10;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 35 + 10;
const int MAXT = 100000 + 10;
using namespace std;
LL a[MAXN];
map<LL, LL> mp1;
map<LL, LL> mp2;
vector<LL> v1;
vector<LL> v2;
LL ans, num;
LL Abs(LL x){
    return x >= 0 ? x : -x;
}
void init(){
    mp1.clear();
    mp2.clear();
    v1.clear();
    v2.clear();
}
void solve(int l, int r, map<LL, LL> &mp, vector<LL> &v){
    int n = r - l + 1;
    for(int i = 1; i < (1 << n); ++i){
        LL sum = 0;
        LL cnt = 0;
        for(int j = 0; j < n; ++j){
            if(i & (1 << j)){
                sum += a[l + j];
                ++cnt;
            }
        }
        if(mp.count(sum))
            mp[sum] = Min(mp[sum], cnt);
        else
            mp[sum] = cnt;
    }
    for(map<LL, LL>::iterator it = mp.begin(); it != mp.end(); ++it){
        v.push_back((*it).first);
        if(Abs((*it).first) < ans){
            ans = Abs((*it).first);
            num = (*it).second;
        }
        else if(ans == Abs((*it).first)){
            num = Min(num, (*it).second);
        }
    }
}
void judge(LL x){
    int l = 0, r = v2.size() - 1;
    while(l <= r){
        int mid = l + (r - l) / 2;
        if(Abs(x + v2[mid]) < ans){
            ans = Abs(x + v2[mid]);
            num = mp1[x] + mp2[v2[mid]];
        }
        else if(Abs(x + v2[mid]) == ans){
            num = Min(num, mp1[x] + mp2[v2[mid]]);
        }
        if(x + v2[mid] < 0) l = mid + 1;
        else r = mid - 1;
    }
}
int main(){
    int N;
    while(scanf("%d", &N) == 1){
        if(!N) return 0;
        init();
        for(int i = 0; i < N; ++i){
            scanf("%lld", &a[i]);
        }
        if(N == 1){
            printf("%lld 1\n", Abs(a[0]));
            continue;
        }
        ans = LL_INF, num = LL_INF;
        solve(0, N / 2 - 1, mp1, v1);
        solve(N / 2, N - 1, mp2, v2);
        int len = v1.size();
        sort(v2.begin(), v2.end());
        for(int i = 0; i < len; ++i){
            judge(v1[i]);
        }
        printf("%lld %lld\n", ans, num);
    }
    return 0;
}