查找網格中的最小未填充索引-有解無憂

為了這個問題，我將所有網格索引、頂點或其他點稱為點。 所以 (1, 1) 是網格上的一個點，例如。

我正在尋找網格中最小的總大小索引。在這種情況下，(1, 1) 的總和為 2，(2, 1) 的總和為 3，對于 (1, 2) 也是如此。

預期結果：

{ X = 1, Y = 1 }
{ X = 2, Y = 1 }
{ X = 1, Y = 2 }
{ X = 2, Y = 2 }
{ X = 3, Y = 1 }
{ X = 3, Y = 2 }
{ X = 1, Y = 3 }
{ X = 2, Y = 3 }
{ X = 3, Y = 3 }

實際結果：

{ X = 1, Y = 1 }
{ X = 2, Y = 1 }
{ X = 1, Y = 2 }
{ X = 2, Y = 2 }
{ X = 3, Y = 2 }
{ X = 2, Y = 3 }
{ X = 3, Y = 3 }
{ X = 4, Y = 3 }
{ X = 3, Y = 4 }

代碼：

local chunks = {}
local previous = {
    X = 0,
    Y = 0
}

local largest = {
    X = 0,
    Y = 0
}

local function addScene()
    local new = {
        X = (previous.X > previous.Y and previous.Y) or previous.X   1,
        Y = (previous.X > previous.Y and previous.X) or previous.Y
    }
    
    if new.X == 0 then
        new.X = 1
    end
    
    if new.Y == 0 then
        new.Y = 1
    end
    
    largest.X = math.max(largest.X, new.X)
    largest.Y = math.max(largest.Y, new.Y)
    
    previous.X = new.X
    previous.Y = new.Y
    
    table.insert(chunks, new)
    return new
end

for i = 1, 3*3, 1 do
    local scene = addScene()
    print(i, "(" .. tostring(scene.X) .. ", " .. tostring(scene.Y) .. ")" .. "\n")
end

為自己測驗

如上所示，此代碼非常接近我希望它執行的操作，但只能達到一定程度，并且必須在越來越大的網格中進行擴展。

uj5u.com熱心網友回復：

盡管很可能效率低下，但這最終會產生接近預期的結果。無論如何，您可以簡單地重新組織表格以使其與我在問題中顯示的完全一樣。

local MAX_SIZE = {
    X = 3,
    Y = 3
}

local chunks = {}
local previous = {
    X = 0,
    Y = 0
}

local function addScene()
    local least = MAX_SIZE.X   MAX_SIZE.Y
    local new = {
        X = 0,
        Y = 0
    }
    
    for x = MAX_SIZE.X, 1, -1 do
        for y = MAX_SIZE.Y, 1, -1 do
            if chunks[tostring(x) .. '-' .. tostring(y)] then
                break
            end
            
            if x ~= 0 and y ~= 0 and x   y <= least then
                least = x   y
                new.X = x
                new.Y = y
            end
        end
    end
    
    if new.X == 0 then
        new.X = 1
    end
    
    if new.Y == 0 then
        new.Y = 1
    end
    
    previous.X = new.X
    previous.Y = new.Y
    chunks[tostring(new.X) .. '-' .. tostring(new.Y)] = new
    return new
end

for i = 1, 3*3, 1 do
    local scene = addScene()
    print(i, "(" .. tostring(scene.X) .. ", " .. tostring(scene.Y) .. ")" .. "\n")
end

試試看

結果：

1 (1, 1)
2 (1, 2)
3 (2, 1)
4 (1, 3)
5 (2, 2)
6 (3, 1)
7 (2, 3)
8 (3, 2)
9 (3, 3)

轉載請註明出處，本文鏈接：https://www.uj5u.com/gongcheng/520219.html

標籤：数学lua网格可扩展性最低限度

上一篇：如何計算BST中的預期葉子數？

下一篇：找一個垂直于直線中間的點LineRenderer