John is going on a fishing trip. He has h hours available (1 <= h <= 16), and there are n lakes in the area (2 <= n <= 25) all reachable along a single, one-way road. John starts at lake 1, but he can finish at any lake he wants. He can only travel from one lake to the next one, but he does not have to stop at any lake unless he wishes to. For each i = 1,...,n - 1, the number of 5-minute intervals it takes to travel from lake i to lake i + 1 is denoted ti (0 < ti <=192). For example, t3 = 4 means that it takes 20 minutes to travel from lake 3 to lake 4. To help plan his fishing trip, John has gathered some information about the lakes. For each lake i, the number of fish expected to be caught in the initial 5 minutes, denoted fi( fi >= 0 ), is known. Each 5 minutes of fishing decreases the number of fish expected to be caught in the next 5-minute interval by a constant rate of di (di >= 0). If the number of fish expected to be caught in an interval is less than or equal to di , there will be no more fish left in the lake in the next interval. To simplify the planning, John assumes that no one else will be fishing at the lakes to affect the number of fish he expects to catch. Write a program to help John plan his fishing trip to maximize the number of fish expected to be caught. The number of minutes spent at each lake must be a multiple of 5.这道题的分析

这道题是一道动态规划问题，需要考虑到以下几个因素：

1. 状态定义：设f(i, j)表示从起点到i时，已经用了j个5分钟的时间，能抓到的最大鱼数。

2. 状态转移：因为从i到j只能走一次，所以可以考虑枚举上一个节点k，转移方程为：f(i, j) = max{f(k, j-ti) + max(0, fi-(j-ti)*di)}，其中ti表示从节点k到节点i需要的时间，fi表示节点i初始时能抓到的鱼数，di表示每5分钟减少的鱼数。

3. 边界条件：f(1, 0) = f(1, 5) = ... = f(1, h*12) = f(1, h*12+1) = ... = f(1, h*12+4) = 0，表示从起点出发，初始时不能抓到任何鱼。

4. 最终答案：最终答案为max{f(i, j)}，其中i表示所有节点中的任意一个节点，j表示用的时间不超过h小时。

需要注意的是，因为每个节点只能经过一次，所以在状态转移时需要枚举上一个节点k，而不能枚举用了多长时间。而且因为时间只能是5的倍数，所以需要将h换算成12h。

相关问题

怎么精简代码func BasinTree(id string) ([]*models.Basin, error) { var basins []*models.Basin res := common.DB.Where("watershed_id = ?", id).Find(&basins) for _, item := range basins { if res.RowsAffected > 0 { //查询流域内所有河道 var subrivers []*models.SubRiver var rivers models.PsRiver common.DB.Model(&rivers).Where("watershed_id = ?", item.ID).Find(&subrivers) item.SubRivers = subrivers var totalL float64 common.DB.Table("ps_rivers").Select("COALESCE(sum(segment_length), 0)").Where("watershed_id = ?", item.ID).Scan(&totalL) item.TotalLength = totalL //查询流域内所有湖泊 var sublakes []*models.SubLake var lakes models.PsLake common.DB.Model(&lakes).Where("watershed_id = ?", item.ID).Find(&sublakes) var totalA float64 common.DB.Table("ps_lakes").Select("COALESCE(sum(area),0)").Where("watershed_id = ?", item.ID).Scan(&totalA) item.TotalArea = totalA item.SubLakes = sublakes } } for _, item := range basins { if res.RowsAffected > 0 { id = strconv.FormatUint(uint64(item.ID), 10) item.SubBasins, _ = BasinTree(id) for _, v := range item.SubBasins { item.TotalArea = item.TotalArea + v.TotalArea item.TotalLength = item.TotalLength + v.TotalLength } if len(item.SubBasins) == 0 { return nil, nil } } } return basins, nil } func BasinInfo(ctx *gin.Context) { id := ctx.Query("id") var req models.Basin var err error resp := models.Response{ Code: 0, Msg: "success", } if len(id) == 0 { resp.Code = 400 resp.Msg = "请输入id值" ctx.JSON(400, resp) return } res := common.DB.Where("id = ?", id).Take(&req) if res.Error != nil { resp.Code = 400 resp.Msg = "查询失败" resp.Data = res.Error ctx.JSON(400, resp) return } //查询流域内所有河道 var subrivers []*models.SubRiver var rivers models.PsRiver var totalL float64 common.DB.Model(&rivers).Where("watershed_id = ?", id).Find(&subrivers) common.DB.Table("ps_rivers").Select("COALESCE(sum(segment_length), 0)").Where("watershed_id = ?", id).Scan(&totalL) req.SubRivers = subrivers req.TotalLength = totalL //查询流域内所有湖泊 var sublakes []*models.SubLake var lakes models.PsLake var totalA float64 common.DB.Model(&lakes).Where("watershed_id = ?", id).Find(&sublakes) common.DB.Table("ps_lakes").Select("COALESCE(sum(area),0)").Where("watershed_id = ?", id).Scan(&totalA) req.SubLakes = sublakes req.TotalArea = totalA req.SubBasins, err = BasinTree(id) if err != nil { resp.Code = 500 resp.Msg = "创建树失败" resp.Data = err ctx.JSON(500, resp) return } for _, v := range req.SubBasins { req.TotalArea = req.TotalArea + v.TotalArea req.TotalLength = req.TotalLength + v.TotalLength } resp.Data = req ctx.JSON(200, resp) }

可以采用递归的方式，将查询流域信息以及子流域信息的代码进行整合，避免代码重复。

func getSubRivers(id uint) ([]*models.SubRiver, float64) {
	var subrivers []*models.SubRiver
	var rivers models.PsRiver
	var totalL float64
	common.DB.Model(&rivers).Where("watershed_id = ?", id).Find(&subrivers)
	common.DB.Table("ps_rivers").Select("COALESCE(sum(segment_length), 0)").Where("watershed_id = ?", id).Scan(&totalL)
	return subrivers, totalL
}

func getSubLakes(id uint) ([]*models.SubLake, float64) {
	var sublakes []*models.SubLake
	var lakes models.PsLake
	var totalA float64
	common.DB.Model(&lakes).Where("watershed_id = ?", id).Find(&sublakes)
	common.DB.Table("ps_lakes").Select("COALESCE(sum(area),0)").Where("watershed_id = ?", id).Scan(&totalA)
	return sublakes, totalA
}

func BasinTree(id string) ([]*models.Basin, error) {
	var basins []*models.Basin
	res := common.DB.Where("watershed_id = ?", id).Find(&basins)
	if res.RowsAffected == 0 {
		return nil, nil
	}
	for _, item := range basins {
		item.SubRivers, item.TotalLength = getSubRivers(item.ID)
		item.SubLakes, item.TotalArea = getSubLakes(item.ID)
		item.SubBasins, _ = BasinTree(strconv.FormatUint(uint64(item.ID), 10))
		for _, v := range item.SubBasins {
			item.TotalArea += v.TotalArea
			item.TotalLength += v.TotalLength
		}
	}
	return basins, nil
}

func BasinInfo(ctx *gin.Context) {
	id := ctx.Query("id")
	var req models.Basin
	var err error
	resp := models.Response{
		Code: 0,
		Msg:  "success",
	}
	if len(id) == 0 {
		resp.Code = 400
		resp.Msg = "请输入id值"
		ctx.JSON(400, resp)
		return
	}
	res := common.DB.Where("id = ?", id).Take(&req)
	if res.Error != nil {
		resp.Code = 400
		resp.Msg = "查询失败"
		resp.Data = res.Error
		ctx.JSON(400, resp)
		return
	}
	req.SubRivers, req.TotalLength = getSubRivers(req.ID)
	req.SubLakes, req.TotalArea = getSubLakes(req.ID)
	req.SubBasins, err = BasinTree(id)
	if err != nil {
		resp.Code = 500
		resp.Msg = "创建树失败"
		resp.Data = err
		ctx.JSON(500, resp)
		return
	}
	for _, v := range req.SubBasins {
		req.TotalArea += v.TotalArea
		req.TotalLength += v.TotalLength
	}
	resp.Data = req
	ctx.JSON(200, resp)
}

Develop and analyze a mathematical model that will assist negotiators to respond to a fixed set of water supply and demand conditions. Use the model to inform dam operations: When the water level in Lake Mead is M and the water level in Lake Powell is P, how much water should be drawn from each lake to meet stated demands? If no additional water is supplied (from rainfall, etc.), and considering the demands as fixed, how long will it take before the demands are not met? How much additional water must be supplied over time to ensure that these fixed demands are met?

Mathematical models can be useful tools in analyzing water supply and demand conditions and making decisions about dam operations. One approach to modeling this situation is to use a linear programming model that maximizes the amount of water supplied while meeting the fixed demand constraints. The objective function is to maximize the total amount of water supplied, and the constraints are the demand requirements and the availability of water in Lake Mead and Lake Powell.

Given the water level in Lake Mead (M) and Lake Powell (P), let x and y be the amounts of water drawn from Lake Mead and Lake Powell, respectively. The objective function can be written as:

Maximize: x + y

Subject to:

x + y >= D (where D is the total fixed demand for water)
x <= M (the amount of water available in Lake Mead)
y <= P (the amount of water available in Lake Powell)

x, y >= 0 (non-negativity constraints)

The solution to this linear programming model will give the optimal amounts of water to be drawn from each lake to meet the fixed demand.

If no additional water is supplied and the demands are fixed, it will take until the water in either Lake Mead or Lake Powell is depleted, or both, until the demands are not met. At that point, the optimization model will not have a feasible solution, and additional water must be supplied to meet the demand.

To determine the amount of additional water that must be supplied over time, the water levels in both lakes must be modeled over time, including inputs such as rainfall and evaporation. The model can then be used to determine the minimum amount of additional water that must be supplied each year to ensure that the fixed demands are met in the future.

This is just one possible approach to modeling this situation, and there may be other models that could also be used to assist negotiators in responding to water supply and demand conditions.