fs25-adsmartpickup/FS25_ADSmartPickup/adChainModel.lua

-- AD Chain Model (ren logik)
-- Sporer baglæns fra de produktioner spilleren selv har tændt (rødder) gennem gårdens egne fabrikker og stalde
-- til råvarerne. Alle mængder er liter pr. måned; varer og afgrøder er NAVNE. Se spec 2026-09-30-produktionskaeder.

ADChainModel = {}
ADChainModel.MAX_DEPTH = 8

local function byKey(a, b) return tostring(a.key) < tostring(b.key) end

-- producenter pr. vare: tændte først, så stabil rækkefølge
local function indexProducers(recipes)
    local index = {}
    for _, recipe in ipairs(recipes or {}) do
        for name, amount in pairs(recipe.outputs or {}) do
            if (amount or 0) > 0 and (recipe.cyclesPerMonth or 0) > 0 then
                index[name] = index[name] or {}
                table.insert(index[name], recipe)
            end
        end
    end
    for _, list in pairs(index) do
        table.sort(list, function(a, b)
            if (a.enabled == true) ~= (b.enabled == true) then return a.enabled == true end
            return byKey(a, b)
        end)
    end
    return index
end

local function indexHerds(husbandry)
    local index = {}
    for _, herd in ipairs(husbandry or {}) do
        for name, perMonth in pairs(herd.outputs or {}) do
            if (perMonth or 0) > 0 then
                index[name] = index[name] or {}
                table.insert(index[name], {building = herd.building, buildingName = herd.buildingName, left = perMonth})
            end
        end
    end
    return index
end

local function sortedNames(map)
    local names = {}
    for name in pairs(map or {}) do table.insert(names, name) end
    table.sort(names)
    return names
end

-- fieldGoods = {[vare] = true}: varer der høstes/laves på marken uden at være en afgrødes hovedvare (hø, hakkelse,
-- halm …). Uden producent/stald er kilden "field" uden afgrøde: intet markbehov og intet problem.
function ADChainModel.build(recipes, husbandry, crops, fieldGoods)
    local chain = {roots = {}, demand = {}, fieldDemand = {}, needRecipes = {}, problems = {}}
    local producers, herds = indexProducers(recipes), indexHerds(husbandry)
    -- Rødderne kører for sig selv med fuld fart: deres egne input kræves af deres eget træ. Som leverandør til en
    -- anden rod har de derfor fuld kapacitet, men deres input regnes ikke med igen (standalone = ingen rekursion).
    -- En pauset rod (recipe.paused) kører også for sig selv, men er ikke rod: kæden skaber intet nyt for den.
    local capacity, roots, standalone = {}, {}, {}
    for _, recipe in ipairs(recipes or {}) do
        capacity[recipe.key] = recipe.cyclesPerMonth or 0
        if recipe.enabled and not recipe.byChain then
            if not recipe.paused then table.insert(roots, recipe) end
            standalone[recipe.key] = true
        end
    end
    table.sort(roots, byKey)

    -- detail = manglende liter/md (flaskehals/køb), need = varens behov pr. måned i den gren
    local function problem(kind, fillType, rootKey, detail, need)
        table.insert(chain.problems, {kind = kind, fillType = fillType, rootKey = rootKey, detail = detail, need = need})
    end

    local need
    local function inputsOf(recipe, cycles, depth, path, rootKey)
        local children = {}
        for _, name in ipairs(sortedNames(recipe.inputs)) do
            local amount = recipe.inputs[name] or 0
            if amount > 0 and cycles > 0 then
                local child = need(name, amount * cycles, depth + 1, path, rootKey)
                child.consumer, child.consumerName = recipe.building, recipe.buildingName
                table.insert(children, child)
            end
        end
        return children
    end

    need = function(name, perMonth, depth, path, rootKey)
        local node = {fillType = name, perMonth = perMonth, coveredPerMonth = 0, producers = {}, herds = {}, children = {}}
        chain.demand[name] = (chain.demand[name] or 0) + perMonth
        if depth > ADChainModel.MAX_DEPTH or path[name] then
            node.source = "cycle"
            problem("cycle", name, rootKey, nil, perMonth)
            return node
        end
        path[name] = true
        local rest = perMonth
        -- a) gårdens fabrikker
        local makers = producers[name]
        if makers ~= nil then
            node.source = "production"
            for _, recipe in ipairs(makers) do
                local perCycle = recipe.outputs[name]
                local cycles = math.min(rest / perCycle, capacity[recipe.key] or 0)
                if cycles > 0 then
                    capacity[recipe.key] = capacity[recipe.key] - cycles
                    rest = rest - cycles * perCycle
                    table.insert(node.producers, {key = recipe.key, building = recipe.building, buildingName = recipe.buildingName, cycles = cycles})
                    if not standalone[recipe.key] then
                        chain.needRecipes[recipe.key] = (chain.needRecipes[recipe.key] or 0) + cycles
                        for _, child in ipairs(inputsOf(recipe, cycles, depth, path, rootKey)) do table.insert(node.children, child) end
                    end
                end
                if rest <= 1e-9 then break end
            end
            if rest > 1e-9 and #node.producers > 0 then
                problem("bottleneck", name, rootKey, rest, perMonth)
                node.shortPerMonth = rest
            end
            if rest > 1e-9 and #node.producers == 0 then
                node.source = nil
            end
        end
        -- b) stalde
        if rest > 1e-9 and #node.producers == 0 and herds[name] ~= nil then
            node.source = "husbandry"
            for _, herd in ipairs(herds[name]) do
                local take = math.min(rest, herd.left)
                if take > 0 then
                    herd.left = herd.left - take
                    rest = rest - take
                    table.insert(node.herds, {building = herd.building, buildingName = herd.buildingName, perMonth = take})
                end
            end
        end
        -- c) mark
        if rest > 1e-9 and #node.producers == 0 and #node.herds == 0 and crops ~= nil and crops[name] ~= nil then
            node.source, node.fruit = "field", crops[name]
            chain.fieldDemand[node.fruit] = (chain.fieldDemand[node.fruit] or 0) + rest
            rest = 0
        end
        -- c2) markvare uden egen afgrøde (fx hø, hakkelse): kommer fra markerne, ingen afgrødeplan
        if rest > 1e-9 and #node.producers == 0 and #node.herds == 0 and fieldGoods ~= nil and fieldGoods[name] then
            node.source, node.fruit = "field", nil
            rest = 0
        end
        -- d) ingen kilde (eller rest efter stald), eller exhausted producers
        if rest > 1e-9 and makers ~= nil and #node.producers == 0 and #node.herds == 0 then
            node.source = "production"
            node.shortPerMonth = rest
            problem("bottleneck", name, rootKey, rest, perMonth)
        elseif rest > 1e-9 and (node.source == nil or node.source == "husbandry") then
            if node.source == nil then node.source = "buy" end
            node.shortPerMonth = rest
            problem("buy", name, rootKey, rest, perMonth)
        end
        node.coveredPerMonth = perMonth - (node.shortPerMonth or 0)
        path[name] = nil
        return node
    end

    for _, root in ipairs(roots) do
        local cycles = root.cyclesPerMonth or 0
        table.insert(chain.roots, {key = root.key, name = root.name, building = root.building, buildingName = root.buildingName,
            children = inputsOf(root, cycles, 0, {}, root.key)})
    end
    return chain
end