我已經有了一點玩這個,這是我迄今爲止,應該給一個很好的起點。我隨機生成了一個黃金和鐵對列表(我用過Point,因爲這對我來說工作起來更簡單,但任何工作都可以)。
這個想法是採取一組小值黃金和交換他們從另一個列表中獲得一個較大的黃金價值。在大多數情況下,會談論相當數量的黃金,但將較大的價值換成較小的黃金。
private void button2_Click(object sender, EventArgs e)
{
var GoldIron = new List<Point>(
new Point[]{
new Point(16,23),new Point(16,28),new Point(19,44),new Point(21,29),
new Point(23,16),new Point(24,82),new Point(27,85),new Point(31,63),
new Point(31,78),new Point(32,65),new Point(41,23),new Point(43,79),
new Point(44,76),new Point(45,23),new Point(47,16),new Point(50,15),
new Point(50,37),new Point(52,28),new Point(52,58),new Point(52,71),
new Point(61,39),new Point(61,75),new Point(63,59),new Point(68,25),
new Point(68,61),new Point(70,24),new Point(71,75),new Point(74,78),
new Point(77,59),new Point(82,27)}
);
listBox1.DataSource = GoldIron;
//Split into 2 lists based on the gold amount
var Left = new List<Point>();
var Right = new List<Point>();
var SumGold = GoldIron.Sum(P => P.X);
var SumIron = GoldIron.Sum(P => P.Y);
label2.Text = SumGold.ToString();
label1.Text = SumIron.ToString();
var LeftGold = 0;
Int32 i = 0;
while (LeftGold < SumGold/2)
{
LeftGold += GoldIron[i].X;
Left.Add(GoldIron[i++]);
}
while (i < GoldIron.Count)
{
Right.Add(GoldIron[i++]);
}
Int32 LIndex = 0;
//Start Algorithm
Int32 LeftIron = Left.Sum(P => P.Y);
Int32 RightIron = Right.Sum(P => P.Y);
while (LeftIron - RightIron > 50 || RightIron - LeftIron > 50)
{
if (LeftIron < RightIron)
{
List<Point> TempList = Left;
Left = Right;
Right = TempList;
LIndex = 0;
}
Int32 SmallestRight = Right[LIndex].X;
LeftGold = 0;
i = 0;
while (LeftGold < SmallestRight)
{
LeftGold += Right[i++].X;
}
Point Temp = Right[LIndex];
Right.RemoveAt(LIndex);
Right.AddRange(Left.Take(i));
Left.RemoveRange(0, i);
Left.Add(Temp);
LIndex += i;
//Sort
Left.Sort(CompareGold);
Right.Sort(CompareGold);
LeftIron = Left.Sum(P => P.Y);
RightIron = Right.Sum(P => P.Y);
}
listBox2.DataSource = Left;
SumGold = Left.Sum(P => P.X);
SumIron = Left.Sum(P => P.Y);
label4.Text = SumGold.ToString();
label3.Text = SumIron.ToString();
listBox3.DataSource = Right;
SumGold = Right.Sum(P => P.X);
SumIron = Right.Sum(P => P.Y);
label6.Text = SumGold.ToString();
label5.Text = SumIron.ToString();
}
而結果: 
你的最後一段聽起來像是你只是想保持你的門限之下的鐵的差異和黃金的量是無關緊要的? – Jan
@Jan不夠完美。黃金銷售絕對應該是公平的,但它並不像保持50以下的鐵差別那麼重要。 編輯:發行區域1:100iron,200gold; area2:50iron,200gold優於: area1:100iron,500gold;區域2:100iron,200gold。 – user1756192