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;
        }
    }
}