題目：

3
75 15 21 
75 15 28 
34 70 5 

188

狀態壓縮 動態規劃

程式碼:

#include<iostream>
using namespace std;

int n,
 
 num[20000], s = 1, maxx[20][20000], number[20];

void build(int k)
{
	if (k >20)return;
	for (int i = 0; num[i] < 1 << (k - 2); i++)num[++s] = num[i] + (1 << (k - 1));
	build(k + 1);
}

int getsum(int j)
{
	int sum = 0;
	for (int i = 0; i < n; i++)sum += ((j&(1 << i))>0)*number[i];
 

	return sum;
}

int main()
{
	num[0] = 0, num[1] = 1;
	build(2);
	while (cin >> n)
	{
		if (n == 0)
		{
			cout << "0\n";
			continue;
		}
		for (int i = 0; i < n; i++)for (int j = 0; num[j] < (1 << n); j++)maxx[i][j] = 0;
		for (int i = 0; i < n; i++)cin >> number[i];
		for
 
 (int j = 0; num[j] < (1 << n); j++)maxx[0][j] = getsum(num[j]);
		for (int i = 1; i < n; i++)
		{
			for (int i = 0; i < n; i++)cin >> number[i];
			for (int j = 0; num[j] < (1 << n); j++)for (int k = 0; num[k] < (1 << n); k++)
			if ((num[j] & num[k]) == 0 && maxx[i][j] < maxx[i - 1][k] + getsum(num[j]))
				maxx[i][j] = maxx[i - 1][k] + getsum(num[j]);
		}
		int ans = 0;
		for (int j = 0; num[j] < (1 << n); j++)if (ans < maxx[n - 1][j])ans = maxx[n - 1][j];
		cout << ans << endl;
	}
	return 0;
}