POJ 3977 题解

题目

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.

输入格式

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

输出格式

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

样例输入

1
10
3
20 100 -100
0

样例输出

10 1
0 2

题解

我自己想的话，除了暴力没有别的办法，看了题解，照着打了一遍代码，才能理解这个解法

具体思想就是，全都暴力会超时，所以分成两半枚举，左半部分枚举完保存，右半部分枚举后在左半部分二分查询。

这算优化暴力吗，dalao们的暴力也和我的暴力不一样。。。

这个代码使用了不少STL也是一个学习STL的机会

代码

#include <algorithm>
#include <cstdio>
#include <map>
using namespace std;
#define abs(x) ((x) >= 0 ? (x) : -(x))
long long arr[40];
int main() {
    int n;
    while (scanf("%d", &n), n) {
        for (int i = 0; i < n; i++) scanf("%lld", &arr[i]);
        pair<long long, long long> ans(abs(arr[0]), 1);
        map<long long, long long> m;
        map<long long, long long>::iterator it;
        int nl = n >> 1, nr = n - (n >> 1);
        for (int i = 1; i < (1 << nl); i++) {
            long long sum = 0, cnt = 0;
            for (int j = 0; j < nl; j++) {
                if ((i >> j) & 1) sum += arr[j], cnt++;
            }
            long long sum1 = abs(sum);
            if (sum1 < ans.first || (sum1 == ans.first && cnt < ans.second)) ans = make_pair(sum1, cnt);
            if (m[sum]) m[sum] = min(m[sum], cnt);
            else m[sum] = cnt;
        }
        for (int i = 1; i < (1 << nr); i++) {
            long long sum = 0, cnt = 0;
            for (int j = 0; j < nr; j++)
                if ((i >> j) & 1) sum += arr[j + nl], cnt++;
            long long sum1 = abs(sum);
            if (sum1 < ans.first || (sum1 == ans.first && cnt < ans.second)) ans = make_pair(sum1, cnt);
            it = m.lower_bound(-sum);
            if (it != m.end()) {
                long long s = abs(it->first + sum), t = it->second + cnt;
                if (s < ans.first || (s == ans.first && t < ans.second))
                    ans = make_pair(s, t);
            }
            if (it != m.begin()) {
                it--;
                long long s = abs(it->first + sum), t = it->second + cnt;
                if (s < ans.first || (s == ans.first && t < ans.second))
                    ans = make_pair(s, t);
            }
        }
        printf("%lld %lld
", ans.first, ans.second);
    }
    return 0;
}