POJ3977 Subset 折半枚舉

    題目大意是給定N個數的集合，從這個集合中找到一個非空子集，使得該子集元素和的絕對值最小，如果有多個答案，輸出元素個數最少的那個。   

N最多爲35，如果直接枚舉顯然是不行的。但是如果我們將這些數分成兩半後再枚舉的話，最多有2^18(262144)，此時我們兩半枚舉後的結果進行排序後再二分搜索一下就可以了。複雜度爲O(nlogn) n最多2^18。

#include <stdio.h>
#include <vector>
#include <math.h>
#include <string.h>
#include <string>
#include <iostream>
#include <queue>
#include <list>
#include <algorithm>
#include <stack>
#include <map>

using namespace std;

struct MyStruct
{
	long long res;
	int i;
};

int compp(const void* a1, const void* a2)
{
	long long dif = ((MyStruct*)a1)->res - ((MyStruct*)a2)->res;
	if (dif > 0)
	{
		return 1;
	}
	else if (dif == 0)
	{
		return 0;
	}
	else
		return -1;
}

MyStruct res[2][300000];

inline long long absll(long long X)
{
	if (X < 0)
	{
		return X * (-1);
	}
	else
		return X;
}

int main()
{
	int n;
#ifdef _DEBUG
	freopen("d:\\in.txt", "r", stdin);
#endif
	long long values[36];
	while (scanf("%d", &n) != EOF)
	{
		if (n == 0)
		{
			break;
		}
		for (int i = 0; i < n; i++)
		{
			scanf("%I64d", &values[i]);
		}
		int maxn = n - n / 2;
		int maxm = n - maxn;
		memset(res, 0, sizeof(res));
		for (int i = 0; i < 1 << maxn; i++)
		{
			res[0][i].i = i;
			for (int k = 0; k < 19; k++)
			{
				if ((i >> k) & 1)
				{
					res[0][i].res += values[k];
				}
			}
		}
		qsort(res[0], 1 << maxn, sizeof(MyStruct), compp);
		for (int i = 0; i < 1 << maxm; i++)
		{
			res[1][i].i = i;
			for (int k = 0; k < 19; k++)
			{
				if ((i >> k) & 1)
				{
					res[1][i].res += values[k + maxn];
				}
			}
		}
		qsort(res[1], 1 << maxm, sizeof(MyStruct), compp);
		long long minvalue = 1000000000000000LL;
		int mink = 32;
		int l = 0;
		int r = (1 << maxm);
		for (int i = 0; i < 1 << maxn; i++)
		{
			l = 0;
			int curk = 0;
			for (int k = 0; k < maxn; k++)
			{
				if ((res[0][i].i >> k) & 1)
				{
					curk++;
				}
			}
			while (r - l > 1)
			{
				int mid = (l + r) / 2;
				long long sum = res[1][mid].res + res[0][i].res;
				if (sum > 0)
				{
					r = mid;
				}
				else
					l = mid;
			}
			
			l = l >= 1 ? l - 1 : l;
			for (int k = l; k < (1 << maxm);k++)
			{	
				int curm = 0;
				for (int m = 0; m < maxm; m++)
				{
					if ((res[1][k].i >> m) & 1)
					{
						curm++;
					}
				}
				if (curm == 0 && curk == 0)
				{
					continue;
				}
				long long sum = res[1][k].res + res[0][i].res;
				if (absll(sum) < minvalue)
				{
					mink = curm + curk;
					minvalue = absll(sum);
				}
				else if (absll(sum) == minvalue)
				{
					mink = min(mink, curk + curm);
				}
				else if (sum > 0)
				{
					break;
				}
			}
		}
		printf("%I64d %d\n", minvalue, mink);
	}
	return 0;
}