153 lines
6.4 KiB
Lua
153 lines
6.4 KiB
Lua
-- AD Chain Model (ren logik)
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-- Sporer baglæns fra de produktioner spilleren selv har tændt (rødder) gennem gårdens egne fabrikker og stalde
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-- til råvarerne. Alle mængder er liter pr. måned; varer og afgrøder er NAVNE. Se spec 2026-09-30-produktionskaeder.
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ADChainModel = {}
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ADChainModel.MAX_DEPTH = 8
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local function byKey(a, b) return tostring(a.key) < tostring(b.key) end
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-- producenter pr. vare: tændte først, så stabil rækkefølge
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local function indexProducers(recipes)
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local index = {}
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for _, recipe in ipairs(recipes or {}) do
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for name, amount in pairs(recipe.outputs or {}) do
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if (amount or 0) > 0 and (recipe.cyclesPerMonth or 0) > 0 then
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index[name] = index[name] or {}
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table.insert(index[name], recipe)
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end
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end
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end
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for _, list in pairs(index) do
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table.sort(list, function(a, b)
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if (a.enabled == true) ~= (b.enabled == true) then return a.enabled == true end
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return byKey(a, b)
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end)
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end
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return index
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end
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local function indexHerds(husbandry)
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local index = {}
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for _, herd in ipairs(husbandry or {}) do
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for name, perMonth in pairs(herd.outputs or {}) do
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if (perMonth or 0) > 0 then
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index[name] = index[name] or {}
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table.insert(index[name], {building = herd.building, buildingName = herd.buildingName, left = perMonth})
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end
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end
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end
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return index
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end
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local function sortedNames(map)
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local names = {}
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for name in pairs(map or {}) do table.insert(names, name) end
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table.sort(names)
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return names
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end
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function ADChainModel.build(recipes, husbandry, crops)
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local chain = {roots = {}, demand = {}, fieldDemand = {}, needRecipes = {}, problems = {}}
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local producers, herds = indexProducers(recipes), indexHerds(husbandry)
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local capacity, roots = {}, {}
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for _, recipe in ipairs(recipes or {}) do
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capacity[recipe.key] = recipe.cyclesPerMonth or 0
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if recipe.enabled and not recipe.byChain then table.insert(roots, recipe) end
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end
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table.sort(roots, byKey)
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for _, root in ipairs(roots) do capacity[root.key] = 0 end -- rødderne kører for sig selv
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local function problem(kind, fillType, rootKey, detail)
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table.insert(chain.problems, {kind = kind, fillType = fillType, rootKey = rootKey, detail = detail})
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end
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local need
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local function inputsOf(recipe, cycles, depth, path, rootKey)
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local children = {}
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for _, name in ipairs(sortedNames(recipe.inputs)) do
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local amount = recipe.inputs[name] or 0
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if amount > 0 and cycles > 0 then
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local child = need(name, amount * cycles, depth + 1, path, rootKey)
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child.consumer, child.consumerName = recipe.building, recipe.buildingName
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table.insert(children, child)
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end
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end
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return children
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end
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need = function(name, perMonth, depth, path, rootKey)
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local node = {fillType = name, perMonth = perMonth, coveredPerMonth = 0, producers = {}, herds = {}, children = {}}
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chain.demand[name] = (chain.demand[name] or 0) + perMonth
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if depth > ADChainModel.MAX_DEPTH or path[name] then
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node.source = "cycle"
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problem("cycle", name, rootKey)
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return node
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end
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path[name] = true
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local rest = perMonth
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-- a) gårdens fabrikker
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local makers = producers[name]
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if makers ~= nil then
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node.source = "production"
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for _, recipe in ipairs(makers) do
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local perCycle = recipe.outputs[name]
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local cycles = math.min(rest / perCycle, capacity[recipe.key] or 0)
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if cycles > 0 then
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capacity[recipe.key] = capacity[recipe.key] - cycles
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rest = rest - cycles * perCycle
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chain.needRecipes[recipe.key] = (chain.needRecipes[recipe.key] or 0) + cycles
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table.insert(node.producers, {key = recipe.key, building = recipe.building, buildingName = recipe.buildingName, cycles = cycles})
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for _, child in ipairs(inputsOf(recipe, cycles, depth, path, rootKey)) do table.insert(node.children, child) end
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end
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if rest <= 1e-9 then break end
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end
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if rest > 1e-9 and #node.producers > 0 then
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problem("bottleneck", name, rootKey, rest)
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node.shortPerMonth = rest
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end
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if rest > 1e-9 and #node.producers == 0 then
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node.source = nil
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end
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end
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-- b) stalde
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if rest > 1e-9 and #node.producers == 0 and herds[name] ~= nil then
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node.source = "husbandry"
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for _, herd in ipairs(herds[name]) do
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local take = math.min(rest, herd.left)
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if take > 0 then
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herd.left = herd.left - take
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rest = rest - take
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table.insert(node.herds, {building = herd.building, buildingName = herd.buildingName, perMonth = take})
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end
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end
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end
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-- c) mark
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if rest > 1e-9 and #node.producers == 0 and #node.herds == 0 and crops ~= nil and crops[name] ~= nil then
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node.source, node.fruit = "field", crops[name]
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chain.fieldDemand[node.fruit] = (chain.fieldDemand[node.fruit] or 0) + rest
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rest = 0
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end
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-- d) ingen kilde (eller rest efter stald), eller exhausted producers
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if rest > 1e-9 and makers ~= nil and #node.producers == 0 and #node.herds == 0 then
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node.source = "production"
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node.shortPerMonth = rest
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problem("bottleneck", name, rootKey, rest)
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elseif rest > 1e-9 and (node.source == nil or node.source == "husbandry") then
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if node.source == nil then node.source = "buy" end
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node.shortPerMonth = rest
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problem("buy", name, rootKey, rest)
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end
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node.coveredPerMonth = perMonth - (node.shortPerMonth or 0)
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path[name] = nil
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return node
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end
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for _, root in ipairs(roots) do
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local cycles = root.cyclesPerMonth or 0
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table.insert(chain.roots, {key = root.key, name = root.name, building = root.building, buildingName = root.buildingName,
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children = inputsOf(root, cycles, 0, {}, root.key)})
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end
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return chain
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end
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