扫雷排雷：数字会告诉你雷在哪

if hiddenCount == grid[r][c] - flagCount then FlagThemAll(hidden) end

fx-mines

local Mines = {}
local KEY = { MINES = 12 }

local function Around(c, r)
    local out = {}
    for dr = -1, 1 do
        for dc = -1, 1 do
            if dr ~= 0 or dc ~= 0 then
                local cc, rr = c + dc, r + dr
                if cc >= 1 and rr >= 1 and cc <= SL:GetValue("cols")
                   and rr <= SL:GetValue("rows") then
                    out[#out + 1] = { c = cc, r = rr }
                end
            end
        end
    end
    return out
end

-- 推理一: 数字 == 隐藏数 - 旗数 -> 剩余全插旗
function Mines.StepFlag()
    for r = 1, SL:GetValue("rows") do
        for c = 1, SL:GetValue("cols") do
            if SL:GetValue("revealed", c, r) and SL:GetValue("num", c, r) > 0 then
                local hidden, flags = 0, 0
                local hiddenOne = nil
                for _, p in ipairs(Around(c, r)) do
                    if SL:GetValue("flagged", p.c, p.r) then
                        flags = flags + 1
                    elseif not SL:GetValue("revealed", p.c, p.r) then
                        hidden = hidden + 1
                        hiddenOne = p
                    end
                end
                if hidden > 0 and hidden == SL:GetValue("num", c, r) - flags then
                    SL:PlaceFlag(hiddenOne.c, hiddenOne.r)
                    return true
                end
            end
        end
    end
    return false
end

-- 推理二: 旗数 == 数字 -> 其余全翻开
function Mines.StepOpen()
    -- 结构同上, 条件换成 flags == num 后 Open(hiddenOne)
end

-- 推理三: 兜底翻一个未标记且非雷的安全格(演示盘保证存在)

return Mines