20231117上机编程[高可靠在线视频]
某电信公司推出高可靠的在线视频业务。为了保证可靠性,公司针对不同视频类型,准备了不同的专用网络通道,并对指定视频类型服务进行通道分配。一个用户在一个时段只能使用一个视频服务,可以多次申请。请实现以下功能:
VideoService(int[] channels, int[] charge)
:初始化系统channels[i]
表示视频类型为i的初始可用通道个数。charge[i]
表示视频类型为i的单位时间费用。- 视频类型仅为:标清视频、高清视频、超清视频,分别用 0、1、2 表示,且对应的带宽从低到高。
allocateChannel(int time, int userId, int videoType)
:在time
时刻,用户userId
申请使用videoType
类型的视频服务。注:计费依照该次申请的视频类型,而非实际占用通道的视频类型。- 如果当前视频类型
videoType
有可用通道,则占用一个通道,并返回 true; - 否则,依次寻找更高带宽的通道,如果有可用,则占用该类型的一个通道,并返回 true;如果仍找不到可用通道,则返回 false 。
freeChannel(int time, int userId)
:在time
时刻,用户userId
申请停止在使用的视频服务:- 如果
userId
正在使用视频服务,则停止该服务并释放所占用的通道,并返回该次服务的费用(可能为 0);否则,返回 -1 。 - 当有通道释放时,如果之前有其他用户因为申请时通道不足而占用了更高带宽的通道,且占用带宽高于释放通道的带宽、所申请带宽不高于释放通道的带宽,则选择一个用户进行降级使用该释放通道。
- 优先选择占用带宽最高的用户;如果还有多个,则选择其中 userId 最小的。
- 注意:降级所释放通道需要重复上述过程。
- 如果
- 如果当前视频类型
queryChannel(int userId)
:查询用户userId
正在使用服务所占用通道的视频类型;若用户不存在或已经停服,则返回-1
。
注:用例保证,1)接口调用中的 time 递增; 2)同一用户停止当前所使用的视频服务后,才会再次申请。
输入
每行表示一个函数调用,初始化函数仅首行调用一次,累计函数调用不超过1000次。
l 1 <= channels[i] <= 1000, 1 <= charge[i] <= 100, channels.length == charge.length == 3
l 1 <= time <= 1000, 1 <= userId <= 1000
l 输入保证videoType
的取值仅为 0、1、2
输出
答题时按照函数/方法原型中的要求(含返回值)进行实现,输出由框架完成(其中首行固定输出 null)
样例1
输入:
VideoService([8, 1, 1], [10, 15, 30])
allocateChannel(3, 107, 1)
allocateChannel(3, 108, 1)
allocateChannel(5, 110, 1)
queryChannel(108)
freeChannel(13, 108)
输出:
null
true
true
false
2
150
解释:
VideoService([8, 1, 1], [10, 15, 30])
:标清视频、高清视频、超清视频的初始可用通道个数分别为 8、1、1;单位时间费用分别为10、15、30。allocateChannel(3, 107, 1)
:时刻3,用户107申请高清视频服务,返回true。allocateChannel(3, 108, 1)
:时刻3,用户108申请高清视频服务,由于高清视频的可用通道已全被占用,但超清视频还有通道可用,因此该用户实际占用超清通道,返回true。allocateChannel(5, 110, 1)
:时刻5,用户110申请高清视频服务,由于高清视频和超清视频的可用通道已全被占用,返回false。queryChannel(108)
:查询用户108正在使用服务所占用通道的视频类型,返回2(即超清视频)。freeChannel(13, 108)
:时刻13,用户108停止使用视频服务,该次使用时长为10,返回计费为15*10=150。注意虽然该用户占用超清通道,但计费时仍按照其申请的高清视频类型计费。
样例2
输入:
VideoService([1, 1, 2], [5, 10, 20])
allocateChannel(1, 100, 0)
allocateChannel(1, 101, 0)
allocateChannel(2, 102, 0)
allocateChannel(4, 103, 0)
freeChannel(6, 100)
queryChannel(102)
freeChannel(7, 103)
allocateChannel(7, 104, 1)
freeChannel(8, 102)
queryChannel(104)
输出:
null
true
true
true
true
25
0
15
true
30
1
解释:
最初4个allocateChannel操作,4个用户都申请标清视频服务,但用户101占用高清通道。用户102和用户103占用超清通道。freeChannel(6, 100)
:时刻6,用户100停止服务,释放出标清通道。此时101、102、103三个用户都可以降到标清通道。根据规则,优先选择占用带宽最高的用户102和103,再从这两个用户中优先选择 userId 最小的用户102降到标清通道。freeChannel(8, 102)
:时刻8,用户102停止服务,释放出标清通道;用户101降到标清通道,又释放出高清通道;然后用户104从超清通道降到用户101刚释放的高清通道。
题目理解:
这道题需要设计一个视频服务系统,为用户提供视频服务,并且需要分配和释放通道。
1) 视频类型有三种类型:标清、高清、超清,申请时如果低清晰度可用通道不足,可占用更高带宽的可用通道,只用付申请类型的费用。
2) 在释放通道x后,其他用户因为申请通道 m 不足而占用了更高带宽的通道n, 若满足 n > x && m <= x 则需要降级,而且有可能触发连锁降级。
以样例2为例,连锁降级如下图所示:
时刻8触发了连锁降级。注意:时刻8没有首先把104降级的原因是,104申请的视频类型是1,比102释放的0高;等到102释放出1后,104申请的类型和释放的相等,就可以降级到1了。
思路解析:
1) 设计建模
系统的初始通道数量和单位时间费用和具体用户无关,存在VideoService中即可。
每个用户的申请信息,可以单独创建一个User类(可维护性更好),也可以全部记录在VideoService中。
2) 连锁降级的实现方法
有两种方法:
l 递归降级:递归结束条件是找不到需要降级的用户,或者用户释放的占用通道已经是超清通道
l 循环降级:循环退出条件是找不到需要降级的用户,或者用户释放的占用通道已经是超清通道
给的参考1:
package main import ( "fmt" ) type VideoService struct { channels []int charge []int used map[int][]int } func NewVideoService(channels []int, charge []int) *VideoService { return &VideoService{ channels: channels, charge: charge, used: make(map[int][]int), } } func (vs *VideoService) AllocateChannel(time, userID, videoType int) bool { for i := videoType; i < len(vs.channels); i++ { if vs.channels[i] == 0 { continue } vs.used[userID] = []int{i, time, videoType, userID} vs.channels[i]-- return true } return false } func (vs *VideoService) FreeChannel(time, userID int) int { if arr, ok := vs.used[userID]; ok { delete(vs.used, userID) vs.channels[arr[0]]++ videoType := arr[0] for { id := vs.getUserID(videoType, vs.used) if id == -1 { break } nextVideoType := vs.used[id][0] vs.channels[nextVideoType]++ vs.channels[videoType]-- vs.used[id][0] = videoType videoType = nextVideoType } return vs.charge[arr[2]] * (time - arr[1]) } return -1 } func (vs *VideoService) getUserID(videoType int, values map[int][]int) int { for _, arr := range values { if arr[0] > arr[2] && arr[2] <= videoType && arr[0] > videoType { return arr[3] } } return -1 } func (vs *VideoService) QueryChannel(userID int) int { if arr, ok := vs.used[userID]; ok { return arr[0] } return -1 } func main() { channels := []int{10, 10, 10} charge := []int{1, 2, 3} videoService := NewVideoService(channels, charge) fmt.Println(videoService.AllocateChannel(1, 1, 0)) // true fmt.Println(videoService.AllocateChannel(2, 2, 1)) // true fmt.Println(videoService.AllocateChannel(3, 3, 2)) // true fmt.Println(videoService.AllocateChannel(4, 4, 0)) // false fmt.Println(videoService.QueryChannel(1)) // 0 fmt.Println(videoService.QueryChannel(2)) // 1 fmt.Println(videoService.QueryChannel(3)) // 2 fmt.Println(videoService.QueryChannel(4)) // -1 fmt.Println(videoService.FreeChannel(5, 2)) // 6 fmt.Println(videoService.FreeChannel(6, 3)) // 9 fmt.Println(videoService.FreeChannel(7, 10)) // -1 fmt.Println(videoService.QueryChannel(2)) // -1 fmt.Println(videoService.QueryChannel(3)) // -1 }
参考2:
package main import ( "fmt" "sort" ) type User struct { userId int startTime int targetChannel int realChannel int } type VideoService struct { channels []struct { availChannelCount int charge int } users []User } func NewVideoService(channels []int, charge []int) *VideoService { vs := &VideoService{ channels: make([]struct { availChannelCount int charge int }, len(channels)), users: []User{}, } for i := 0; i < len(channels); i++ { vs.channels[i].availChannelCount = channels[i] vs.channels[i].charge = charge[i] } return vs } func (vs *VideoService) AllocateChannel(time, userId, videoType int) bool { for i := videoType; i <= 2; i++ { if vs.channels[i].availChannelCount > 0 { vs.channels[i].availChannelCount-- vs.users = append(vs.users, User{ userId: userId, startTime: time, targetChannel: videoType, realChannel: i, }) return true } } return false } func (vs *VideoService) FreeChannel(time, userId int) int { index := -1 for i, user := range vs.users { if user.userId == userId { index = i break } } if index == -1 { return -1 } freeUser := vs.users[index] vs.users = append(vs.users[:index], vs.users[index+1:]...) totalCharge := (time - freeUser.startTime) * vs.channels[freeUser.targetChannel].charge vs.downGradeChannel(freeUser.realChannel) return totalCharge } func (vs *VideoService) downGradeChannel(videoType int) { upgradedUsers := []User{} for _, user := range vs.users { if user.realChannel > videoType && user.targetChannel <= videoType { upgradedUsers = append(upgradedUsers, user) } } if len(upgradedUsers) == 0 { vs.channels[videoType].availChannelCount++ } else { sort.Slice(upgradedUsers, func(i, j int) bool { if upgradedUsers[i].realChannel > upgradedUsers[j].realChannel { return true } else if upgradedUsers[i].realChannel < upgradedUsers[j].realChannel { return false } else { return upgradedUsers[i].userId < upgradedUsers[j].userId } }) downgradeUser := upgradedUsers[0] typ := downgradeUser.realChannel downgradeUser.realChannel = videoType vs.downGradeChannel(typ) } } func (vs *VideoService) QueryChannel(userId int) int { for _, user := range vs.users { if user.userId == userId { return user.realChannel } } return -1 } func main() { channels := []int{10, 10, 10} charge := []int{1, 2, 3} videoService := NewVideoService(channels, charge) fmt.Println(videoService.AllocateChannel(1, 1, 0)) // true fmt.Println(videoService.AllocateChannel(2, 2, 1)) // true fmt.Println(videoService.AllocateChannel(3, 3, 2)) // true fmt.Println(videoService.AllocateChannel(4, 4, 0)) // false fmt.Println(videoService.QueryChannel(1)) // 0 fmt.Println(videoService.QueryChannel(2)) // 1 fmt.Println(videoService.QueryChannel(3)) // 2 fmt.Println(videoService.QueryChannel(4)) // -1 fmt.Println(videoService.FreeChannel(5, 2)) // 6 fmt.Println(videoService.FreeChannel(6, 3)) // 9 fmt.Println(videoService.FreeChannel(7, 10)) // -1 fmt.Println(videoService.QueryChannel(2)) // -1 fmt.Println(videoService.QueryChannel(3)) // -1 }
待补充的
// 答题框内的代码仅为待实现代码,执行或提交代码时,仅包含待实现部分,不要包含其它代码(也不要包含 package main)。 // 判题时CodeCheck/Cmetrics等工具也仅扫描待实现部分。 // 若需要完整框架用于本地调试,请点击答题框上面的“下载完整框架代码”进行下载。 type videoService struct { channels []int charges []int oldUse map[int][]int realUse map[int]int } func construct(channels []int, charge []int) *videoService { return &videoService{ channels, charge, make(map[int][]int), make(map[int]int), } } func (sys *videoService) allocateChannel(time, userId, videoType int) bool { applyChan := make([]int, 2) applyChan[0] = videoType applyChan[1] = time sys.oldUse[userId] = applyChan for i:=videoType; i<=2; i++ { chanCount := sys.channels[i] if chanCount > 0 { sys.realUse[userId] = i sys.channels[i] -- return true } } return false } func (sys *videoService) freeChannel(time, userId int) int { oldApply, ok := sys.oldUse[userId] if !ok { return -1 } oldChan := oldApply[0] oldTime := oldApply[1] oldCharge := sys.charges[oldChan] realApplyChan, _ := sys.realUse[userId] ansCharge := (time - oldTime) * oldCharge sys.channels[realApplyChan] ++ fmt.Println(sys) delete(sys.oldUse, userId) delete(sys.realUse, userId) // 循环降级 申请值 return ansCharge } func (sys *videoService) queryChannel(userId int) int { realApply,ok := sys.realUse[userId] if !ok { return -1 } fmt.Println(sys) return realApply }
输入: VideoService([1, 1, 1], [50, 60, 70]) allocateChannel(12, 6, 0) allocateChannel(30, 90, 1) allocateChannel(410, 890, 1) freeChannel(500, 6) queryChannel(890) freeChannel(500, 90) queryChannel(890) 输出: null true true true 24400 2 28200 1
总结反思:
// 设计题解题思路: 1. 如果一开头没有设计数据结构的思路,建议先看业务步骤,尝试写每个功能,从功能需要上明确数据结构。 2. 仔细审题,申请占用的缓存条件,降级怎么降,仔细看条件,想思路后再写代码。 type videoService struct { channels []int charges []int oldUse map[int][]int realUse map[int]int } func construct(channels []int, charge []int) *videoService { return &videoService{ channels, charge, make(map[int][]int), make(map[int]int), } } func (sys *videoService) allocateChannel(time, userId, videoType int) bool { applyChan := make([]int, 2) applyChan[0] = videoType applyChan[1] = time sys.oldUse[userId] = applyChan for i:=videoType; i<=2; i++ { chanCount := sys.channels[i] if chanCount > 0 { sys.realUse[userId] = i sys.channels[i] -- return true } } return false } func (sys *videoService) freeChannel(time, userId int) int { //fmt.Println(sys) oldApply, ok := sys.oldUse[userId] if !ok { return -1 } oldChan := oldApply[0] oldTime := oldApply[1] oldCharge := sys.charges[oldChan] realApplyChan, _ := sys.realUse[userId] ansCharge := (time - oldTime) * oldCharge sys.channels[realApplyChan] ++ delete(sys.oldUse, userId) delete(sys.realUse, userId) nowChannel := oldChan // 循环释放通道 当其他的通道不可以降级为止 for nowChannel != -1 { nowChannel = sys.free(nowChannel) } return ansCharge } // 从当前通道删掉了一个人 看看需不需要下降其他用户的通道 需要就返回下降的通道值 否则返回-1 func (sys *videoService) free(channel int) int { // 优先从最高级的通道向下降 for i := 2; i > channel; i-- { // 获取当前遍历的通道 for realUser, realChan := range sys.realUse { if realChan == i { // 默认就是id从小到大排序 for oldUser, oldUseList := range sys.oldUse { oldChan := oldUseList[0] channelCount := sys.channels[channel] //fmt.Println("now chan count", channelCount) //fmt.Println("realChan", realChan, oldChan, i, channel, realUser, oldUser) // 可以降级 if realUser == oldUser && realChan != oldChan && realChan > channel && oldChan <= channel && channelCount > 0 { // 高级通道删除该用户 //fmt.Println("begin del", sys) sys.channels[realChan] ++ // 添加到当前通道 sys.realUse[realUser] = channel sys.channels[channel] -- //fmt.Println("end del", sys) // 通道删掉了一个人 return realChan; } } } } } return -1 } func (sys *videoService) queryChannel(userId int) int { realApply,ok := sys.realUse[userId] if !ok { return -1 } //fmt.Println(sys) return realApply }
pass
// 答题框内的代码仅为待实现代码,执行或提交代码时,仅包含待实现部分,不要包含其它代码(也不要包含 package main)。 // 判题时CodeCheck/Cmetrics等工具也仅扫描待实现部分。 // 若需要完整框架用于本地调试,请点击答题框上面的“下载完整框架代码”进行下载。 import "sort" type User struct { time int id int nowChannel int // 当前占用通道,-1 表示当前没有占用通道 needChannel int // 需要的通道 } type videoService struct { channelMap map[int]*TreeSet userMap map[int]*User channels []int charge []int } type TreeSet struct { set map[*User]bool } func NewTreeSet() *TreeSet { return &TreeSet{ set: make(map[*User]bool), } } func (ts *TreeSet) Add(user *User) { ts.set[user] = true } func (ts *TreeSet) Remove(user *User) { delete(ts.set, user) } func (ts *TreeSet) Size() int { return len(ts.set) } func (ts *TreeSet) Values() []*User { values := make([]*User, 0, len(ts.set)) for user := range ts.set { values = append(values, user) } sort.Slice(values, func(i, j int) bool { return values[i].id < values[j].id }) return values } func construct(channels []int, charge []int) *videoService { channelMap := make(map[int]*TreeSet) userMap := make(map[int]*User) for i := 0; i < 3; i++ { channelMap[i] = NewTreeSet() } return &videoService{ channelMap: channelMap, userMap: userMap, channels: channels, charge: charge, } } func (vs *videoService) getUser(userId int) *User { if _, ok := vs.userMap[userId]; !ok { vs.userMap[userId] = &User{ id: userId, nowChannel: -1, } } return vs.userMap[userId] } func (vs *videoService) allocateChannel(time, userId, videoType int) bool { user := vs.getUser(userId) for i := videoType; i < 3; i++ { channel := vs.channelMap[i] if channel.Size() < vs.channels[i] { channel.Add(user) user.nowChannel = i user.time = time user.needChannel = videoType return true } } return false } func (vs *videoService) free(channel int) int { for i := 2; i > channel; i-- { channelUser := vs.channelMap[i] for _, user := range channelUser.Values() { if user.nowChannel != user.needChannel && user.nowChannel > channel && user.needChannel <= channel { channelUser.Remove(user) vs.channelMap[channel].Add(user) t := user.nowChannel user.nowChannel = channel return t } } } return -1 } func (vs *videoService) freeChannel(time, userId int) int { if _, ok := vs.userMap[userId]; !ok || vs.getUser(userId).nowChannel == -1 { return -1 } user := vs.getUser(userId) res := vs.charge[user.needChannel] * (time - user.time) nowChannel := user.nowChannel user.nowChannel = -1 channel := vs.channelMap[nowChannel] channel.Remove(user) for nowChannel != -1 { nowChannel = vs.free(nowChannel) } return res } func (vs *videoService) queryChannel(userId int) int { if _, ok := vs.userMap[userId]; !ok { return -1 } return vs.getUser(userId).nowChannel }
pass demo
package main import ( "fmt" "sort" ) type User struct { time int id int nowChannel int // 当前占用通道,-1 表示当前没有占用通道 needChannel int // 需要的通道 } type VideoService struct { channelMap map[int]*TreeSet userMap map[int]*User channels []int charge []int } func NewVideoService(channels, charge []int) *VideoService { channelMap := make(map[int]*TreeSet) userMap := make(map[int]*User) for i := 0; i < 3; i++ { channelMap[i] = NewTreeSet() } return &VideoService{ channelMap: channelMap, userMap: userMap, channels: channels, charge: charge, } } func (vs *VideoService) getUser(userId int) *User { if _, ok := vs.userMap[userId]; !ok { vs.userMap[userId] = &User{ id: userId, nowChannel: -1, } } return vs.userMap[userId] } func (vs *VideoService) AllocateChannel(time, userId, videoType int) bool { user := vs.getUser(userId) for i := videoType; i < 3; i++ { channel := vs.channelMap[i] if channel.Size() < vs.channels[i] { channel.Add(user) user.nowChannel = i user.time = time user.needChannel = videoType return true } } return false } func (vs *VideoService) FreeChannel(time, userId int) int { if _, ok := vs.userMap[userId]; !ok || vs.getUser(userId).nowChannel == -1 { return -1 } user := vs.getUser(userId) res := vs.charge[user.needChannel] * (time - user.time) nowChannel := user.nowChannel user.nowChannel = -1 channel := vs.channelMap[nowChannel] channel.Remove(user) for nowChannel != -1 { nowChannel = vs.free(nowChannel) } return res } func (vs *VideoService) free(channel int) int { for i := 2; i > channel; i-- { channelUser := vs.channelMap[i] for _, user := range channelUser.Values() { if user.nowChannel != user.needChannel && user.nowChannel > channel && user.needChannel <= channel { channelUser.Remove(user) vs.channelMap[channel].Add(user) t := user.nowChannel user.nowChannel = channel return t } } } return -1 } func (vs *VideoService) QueryChannel(userId int) int { if _, ok := vs.userMap[userId]; !ok { return -1 } return vs.getUser(userId).nowChannel } type TreeSet struct { set map[*User]bool } func NewTreeSet() *TreeSet { return &TreeSet{ set: make(map[*User]bool), } } func (ts *TreeSet) Add(user *User) { ts.set[user] = true } func (ts *TreeSet) Remove(user *User) { delete(ts.set, user) } func (ts *TreeSet) Size() int { return len(ts.set) } func (ts *TreeSet) Values() []*User { values := make([]*User, 0, len(ts.set)) for user := range ts.set { values = append(values, user) } sort.Slice(values, func(i, j int) bool { return values[i].id < values[j].id }) return values } func main() { channels := []int{3, 3, 3} charge := []int{10, 20, 30} videoService := NewVideoService(channels, charge) videoService.AllocateChannel(1, 1, 0) videoService.AllocateChannel(1, 2, 1) videoService.AllocateChannel(1, 3, 2) fmt.Println(videoService.QueryChannel(1)) // Output: 0 fmt.Println(videoService.QueryChannel(2)) // Output: 1 fmt.Println(videoService.QueryChannel(3)) // Output: 2 fmt.Println(videoService.FreeChannel(5, 2)) // Output: 40 fmt.Println(videoService.QueryChannel(1)) // Output: 0 fmt.Println(videoService.QueryChannel(2)) // Output: -1 fmt.Println(videoService.QueryChannel(3)) // Output: 2 }
本文来自博客园,作者:易先讯,转载请注明原文链接:https://www.cnblogs.com/gongxianjin/p/17860147.html