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
向宇it5 分钟前
【unity组件介绍】URP Decal Projector贴花投影器,将特定材质(贴花)投影到场景中的其他对象上。
游戏·3d·unity·c#·游戏引擎·材质
古月-一个C++方向的小白5 小时前
C++11之lambda表达式与包装器
开发语言·c++
沐知全栈开发6 小时前
Eclipse 生成 jar 包
开发语言
杭州杭州杭州7 小时前
Python笔记
开发语言·笔记·python
tanyongxi667 小时前
C++ AVL树实现详解:平衡二叉搜索树的原理与代码实现
开发语言·c++
Wendy14418 小时前
【线性回归(最小二乘法MSE)】——机器学习
算法·机器学习·线性回归
拾光拾趣录8 小时前
括号生成算法
前端·算法
阿葱(聪)8 小时前
java 在k8s中的部署流程
java·开发语言·docker·kubernetes
渣呵8 小时前
求不重叠区间总和最大值
算法
浮生带你学Java9 小时前
2025Java面试题及答案整理( 2025年 7 月最新版,持续更新)
java·开发语言·数据库·面试·职场和发展