SPOJ11469 SUBSET

题面

Farmer John's owns N cows (2 <= N <= 20), where cow i produces M(i) units of milk each day (1 <= M(i) <= 100,000,000).

FJ wants to streamline the process of milking his cows every day, so he installs a brand new milking machine in his barn.

Unfortunately, the machine turns out to be far too sensitive: it only works properly if the cows on the left side of the

barn have the exact same total milk output as the cows on the right side of the barn!

Let us call a subset of cows "balanced" if it can be partitioned into two groups having equal milk output.

Since only a balanced subset of cows can make the milking machine work, FJ wonders how many subsets of his N cows are balanced.

Please help him compute this quantity.

有多少个非空子集,能划分成和相等的两份。

题解

我在考场上打的是暴力(3^n)，我不会告诉你我CE了

我们可以(3^{n/2})枚举两边的子集，然后(meeting;in;the;middle)即可

代码

#include<cstdio>
#include<map>
#include<vector>
#define RG register
#define clear(x, y) memset(x, y, sizeof(x));

inline int read()
{
	int data = 0, w = 1;
	char ch = getchar();
	while(ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
	if(ch == '-') w = -1, ch = getchar();
	while(ch >= '0' && ch <= '9') data = data * 10 + (ch ^ 48), ch = getchar();
	return data*w;
}

const int maxn(21);
int n, a[maxn], ok[1 << maxn], cnt, ans;

typedef std::vector<int>::iterator iter;
std::map<int, int> map;
std::vector<int> set[1 << maxn];

void dfs(int x, int s, int d)
{
	if(x > (n >> 1) - 1)
	{
		if(map.find(d) == map.end()) map[d] = ++cnt;
		int t = map[d]; set[t].push_back(s); return;
	}

	dfs(x + 1, s, d);
	dfs(x + 1, s | (1 << x), d + a[x]);
	dfs(x + 1, s | (1 << x), d - a[x]);
}

void Dfs(int x, int s, int d)
{
	if(x > n - 1)
	{
		if(map.find(d) == map.end()) return;
		int t = map[d];
		for(RG iter it = set[t].begin(); it != set[t].end(); ++it) ok[(*it) | s] = 1;
		return;
	}

	Dfs(x + 1, s, d);
	Dfs(x + 1, s | (1 << x), d + a[x]);
	Dfs(x + 1, s | (1 << x), d - a[x]);
}

int main()
{
	n = read();
	for(RG int i = 0; i < n; i++) a[i] = read();
	dfs(0, 0, 0); Dfs((n >> 1), 0, 0);
	for(RG int i = (1 << n) - 1; i; i--) ans += ok[i];
	printf("%d
", ans);
	return 0;
}