C#,数值计算——Dynpro的计算方法与源程序

给定向量nstate,其整数值是每个状态中的状态数阶段(第一和最后阶段为1),并给定函数成本(j,k,i)返回在阶段i的状态j和的状态k之间移动的成本阶段i+1,此例程返回与nstate长度相同的向量包含最低成本路径的状态号。状态数从0开始,并且返回向量的第一个和最后一个分量将因此总是0。

Given the vector nstate whose integer values are the number of states in each stage(1 for the first and last stages), and given a function cost(j, k, i)that returns the cost of moving between state j of stage i and state k ofstage i+1, this routine returns a vector of the same length as nstatecontaining the state numbers of the lowest cost path. States number from 0,and the first and last components of the returned vector will thus always be 0.

using System;

namespace Legalsoft.Truffer

{

/// <summary>

/// Given the vector nstate whose integer values are the number of states in each

/// stage(1 for the first and last stages), and given a function cost(j, k, i)

/// that returns the cost of moving between state j of stage i and state k of

/// stage i+1, this routine returns a vector of the same length as nstate

/// containing the state numbers of the lowest cost path. States number from 0,

/// and the first and last components of the returned vector will thus always be 0.

/// </summary>

public abstract class Dynpro

{

public Dynpro()

{

}

public abstract double cost(int jj, int kk, int ii);

public int[] dynpro(int[] nstate)

{

const double BIG = 1.0e99;

const double EPS = float.Epsilon; //numeric_limits<double>.epsilon();

int nstage = nstate.Length - 1;

int[] answer = new int[nstage + 1];

if (nstate[0] != 1 || nstate[nstage] != 1)

{

throw new Exception("One state allowed in first and last stages.");

}

double[,] best = new double[nstage + 1, nstate[0]];

best[0, 0] = 0.0;

for (int i = 1; i <= nstage; i++)

{

for (int k = 0; k < nstate[i]; k++)

{

double b = BIG;

for (int j = 0; j < nstate[i - 1]; j++)

{

double a = best[i - 1, j] + cost(j, k, i - 1);

if ((a) < b)

{

b = a;

}

}

best[i, k] = b;

}

}

answer[nstage] = answer[0] = 0;

for (int i = nstage - 1; i > 0; i--)

{

int k = answer[i + 1];

double b = best[i + 1, k];

int j = 0;

for (; j < nstate[i]; j++)

{

double temp = best[i, j] + cost(j, k, i);

if (Math.Abs(b - temp) <= EPS * Math.Abs(temp))

{

break;

}

}

answer[i] = j;

}

return answer;

}

}

}

相关推荐
苦夏木禾20 分钟前
js请求避免缓存的三种方式
开发语言·javascript·缓存
重庆小透明25 分钟前
力扣刷题记录【1】146.LRU缓存
java·后端·学习·算法·leetcode·缓存
超级土豆粉29 分钟前
Turndown.js: 优雅地将 HTML 转换为 Markdown
开发语言·javascript·html
desssq1 小时前
力扣:70. 爬楼梯
算法·leetcode·职场和发展
金增辉1 小时前
基于C#的OPCServer应用开发,引用WtOPCSvr.dll
c#
clock的时钟1 小时前
暑期数据结构第一天
数据结构·算法
wei_shuo2 小时前
飞算 JavaAI 开发助手:深度学习驱动下的 Java 全链路智能开发新范式
java·开发语言·飞算javaai
熊猫钓鱼>_>2 小时前
用Python解锁图像处理之力:从基础到智能应用的深度探索
开发语言·图像处理·python
小小小小王王王2 小时前
求猪肉价格最大值
数据结构·c++·算法
GO兔2 小时前
开篇:GORM入门——Go语言的ORM王者
开发语言·后端·golang·go