【框架设计】活动排期框架：多活动的时间编排设计

-- 活动排期：配置与校验
local player = class(actor)
local SCHEDULE = {
    {name = "双倍经验", days = {6, 0}, start = 1000, duration = 120, group = "exp"},
    {name = "行会攻城", days = {6}, start = 1400, duration = 120, group = "siege"},
    {name = "押镖双倍", days = {6}, start = 1300, duration = 90, group = "exp"},
}
local function checkScheduleConflicts()
    local conflicts = {}
    for i = 1, #SCHEDULE do
        for j = i + 1, #SCHEDULE do
            local a, b = SCHEDULE[i], SCHEDULE[j]
            if a.group == b.group and a.days[1] == b.days[1] then
                conflicts[#conflicts + 1] = a.name .. " 与 " .. b.name .. " 时段冲突"
            end
        end
    end
    return conflicts
end

-- 活动排期：调度执行
local player = class(actor)
local ACTIVE = {}
local function tickSchedule(player)
    local now = os.date("%H%M")
    local today = tonumber(os.date("%w"))
    for _, item in ipairs(SCHEDULE) do
        local key = item.name .. "_" .. os.date("%Y%m%d")
        if now >= tostring(item.start) and ACTIVE[key] == nil then
            for _, d in ipairs(item.days) do
                if d == today then
                    ACTIVE[key] = true
                    onActivityOpen(player, item.name)
                end
            end
        end
    end
end
function onActivityOpen(player, actName)
    sendcentermsg(player, 255, 0, actName .. " 已开启，快来参与！", 0, 8)
end