yunchou.txt

运筹学算法

 scanf("%d",&flag);
 printf("number of the actions   ");
 scanf("%d",&m);
 printf("number of the nature states    ");
 scanf("%d",&n);
 for(j=0;j<n;j++)
  printf(" t%d",j+1);
 printf(" ");
 for(i=0;i<m;i++)
  {printf("A%d ",i+1);
  for(j=0;j<n;j++)
   scanf("%f",&a[i][j]);
  }
 printf("probability of the natural states=? ");
 for(j=0;j<n;j++)
  scanf("%f",&p[j]);
 if(flag)
  {printf("given data=?   ");
  scanf("%f",&xx);
  printf("states of nature=? ");
  for(j=0;j<n;j++)
   scanf("%d",&state[j]);
  sum=0;
  for(j=0;j<n;j++)
   sum+=p[j]*p(xx,state[j]);
  for(j=0;j<n;j++)
   p[j]=p[j]*p(xx,state[j])/sum;
  }
 for(i=0;i<m;i++)
  {sum=0;
  for(j=0;j<n;j++)
   sum+=p[j]*a[i][j];
  e[i]=sum;
  }
 decision=e[0];
 index=1;
 if(type==0)
  for(i=1;i<m;i++)
  { if(decision<e[i])
    {decision=e[i];
    index=i+1;
    }
  }
 else
  for(i=0;i<m;i++)
   if(decision>e[i])
    {decision=e[i];
    index=i+1;
    }
 printf(" ********** ");
 printf("Results:");
 printf(" ********** ");
 printf("expected loss for each course of action based on prior distribution ");
 for(i=0;i<m;i++)
  printf("%f ",e[i]);
 printf(" The optimum expected loss is %f       ",decision);
 printf(" choose action A( %d )",index);
 return;
}

void main(){
 clrscr();
 decision();
 getch();
}

 （4）dp_invest：动态规划的投资问题
#include <stdio.h>

int istar[10];

float dp_invest(int N,int K){
 int i,j,sum,z,d[10][50];
 float g[10][50],f[10][50];
 printf(" The return function values as follows! ");
 for(j=0;j<K+1;j++)
  printf(" %d",j);
 for(i=0;i<N;i++)
  {printf(" task%d: ",i+1);
  for(j=0;j<K+1;j++)
   scanf("%f",&g[i][j]);
  }
 for(i=0;i<N;i++)
  for(j=0;j<K+1;j++)
   {f[i][j]=0;d[i][j]=0;}
 for(j=0;j<K+1;j++)
  {f[N-1][j]=g[N-1][j];
   d[N-1][j]=j;
   }
 for(i=N-2;i>=0;i--)
  for(j=1;j<K+1;j++)
   {f[i][j]=g[i][0]+f[i+1][j];
   d[i][j]=0;
   for(z=1;z<=j;z++)
    if((g[i][z]+f[i+1][j-z])>f[i][j])
     {f[i][j]=g[i][z]+f[i+1][j-z];
     d[i][j]=z;
     }
   }
 istar[0]=d[0][K];
 for(i=1;i<N;i++)
  {sum=0;
  for(j=0;j<=i-1;j++)
   sum=sum+istar[j];
  istar[i]=d[i][K-sum];
   }
 return f[0][K];
}

void main(){
 int i,N,K;
 clrscr();
 printf("WELCOME TO THE DYNAMIC_INVEST SYSTEM! ");
 printf("How many tasks ?    ");
 scanf("%d",&N);
 printf(" How many units of materials ?    ");
 scanf("%d",&K);
 printf(" The optimal return is            %f",dp_invest(N,K));
 for(i=0;i<N;i++)
  printf(" The optimal amout to invest in task %d  is      %d",i+1,istar[i]);
 getch();
}

 （5）dp_plan：生产计划算法
#include <stdio.h>

#define pc(j) 20+5*j
#define e(j) j

void dp_plan(){
 int i,j,k,sum,limit,n,io,max_stock,max_production,d[10],x[10][100],z,z_maxlimit,z_minlimit,xstar[10];
 float f[10][100],temp;
 printf("WELCOME TO THE DYNIMIC SYSTEM! periods of the inventory =?    ");
 scanf("%d",&n);
 printf(" the maximum stocks of each period=?   ");
 scanf("%d",&max_stock);
 printf(" the maximum production of each period=?   ");
 scanf("%d",&max_production);
 printf(" The stocks of the first period=?   ");
 scanf("%d",&io);
 for(i=0;i<n;i++)
  {printf("Demand for period %d  is    ",i+1);
  scanf("%d",&d[i]);
  }
 for(k=0;k<d[n-1];k++)
  {x[n-1][k]=d[n-1]-k;
  f[n-1][k]=pc(x[n-1][k]);
  }
 f[n-1][d[n-1]]=0;
 x[n-1][d[n-1]]=0;
 for(i=n-2;i>=0;i--)
  {sum=0;
  for(j=i;j<n;j++)
   sum+=d[j];
  if(sum<max_stock)
   limit=sum;
  else
   limit=max_stock;
  for(k=0;k<=limit;k++)
   {if(d[i]-k>0)
    {z_minlimit=d[i]-k;
    f[i][k]=pc(z_minlimit)+e(0)+f[i+1][0];
    x[i][k]=z_minlimit;
    }
   else
    {z_minlimit=0;
    f[i][k]=e(k-d[i])+f[i+1][k-d[i]];
    x[i][k]=0;
    }
   if(sum-k>max_stock+d[i]-k)
    if(max_stock+d[i]-k>max_production)
     z_maxlimit=max_production;
    else
     z_maxlimit=max_stock+d[i]-k;
   else
    if(sum-k>max_production)
     z_maxlimit=max_production;
    else
     z_maxlimit=sum-k;
   for(z=z_minlimit;z<=z_maxlimit;z++)
    {temp=pc(z)+e(k+z-d[i])+f[i+1][k+z-d[i]];
    if(f[i][k]>temp)
     {f[i][k]=temp;
     x[i][k]=z;
     }
    }
   }
  }
/* for(i=0;i<n;i++)
  {printf(" the period %d ",i+1);
  for(j=0;j<=5;j++)
   printf(" %f--->%d ",f[i][j],x[i][j]);
  getch();

  }*/
 printf(" The minimum policy cost for the %d periods is  %f",n,f[0][io]);
 xstar[0]=x[0][io];
 j=io;
 printf(" The optimal amount to produce in period 1 is  %d",xstar[0]);
 for(i=1;i<n;i++)
  {xstar[i]=x[i][xstar[i-1]-d[i-1]+j];
  printf(" The optimal amount to produce in period %d  is  %d",i+1,xstar[i]);
  j=xstar[i-1]-d[i-1]+j;
  }
}

void main(){
 clrscr();
 dp_plan();
 getch();
}

 （6）linear：线性规划单纯形算法
#include <stdio.h>

int K,M,N,Q=100,Type,Get,Let,Et,Code[50],XB[50],IA,IAA[50],Indexg,Indexl,Indexe;
float Sum,A[50][50],B[50],C[50];

void initiate();
void solve();

void main(){
 int i,j;
 clrscr();
  /******   input data   ******/
 printf(" Welcome to the linear programming solution! ******** Notice 1! ******** If the type of objective equation is ?max? ,please enter ?1?! Else please enter ?0?! ");
 printf("******** Notice 2! ******** Define the type of subjective equation as following: ?<=?is equal to ?0?! ?>=? is equal to  ?1?! ?=?is equal to ?2?! ");
 printf(" Please input the coefficients or the constants: ");
 printf("THe number of subjective equations  ?  ");
 scanf("%d",&M);
 printf("THe number of variables ?  ");
 scanf("%d",&K);
 printf("THe number of ? <= ?subjective equations  ?  ");
 scanf("%d",&Let);
 printf("THe number of ? >= ?subjective equations  ?  ");
 scanf("%d",&Get);
 printf("THe number of ? = ?subjective equations  ?  ");
 scanf("%d",&Et);
 printf("THe type of objective equation  ?  ");
 scanf("%d",&Type);
 N=K+Let+Et+2*Get;
  for(i=0;i<M;i++)
  {printf("Please input %d?s equation: ",i+1);
     printf("type ?  ");
    scanf("%d",&Code[i]);
    printf("constant ?  ");
      scanf("%f",&B[i]);
    for(j=0;j<K;j++)
     {printf("coefficient ?  ");
     scanf("%f",&A[i][j]);
          }
        }
   printf("Plese input the constants of the object equation: ");
   for(j=0;j<K;j++)
     scanf("%f",&C[j]);
   printf(" ************************************* Please check the data you just input! *************************************  ");
   getch();
   if(Type)
  printf("The type of the object equation is ?max? ");
 else
  printf("The type of the object equation is ?min? ");
 printf(" The object equation is : ");
 for(j=0;j<K;j++)
  {printf("(%f)X%d ",C[j],j+1);
   if(j!=K-1)
    printf("+");
  }
 printf(" The suject equation is :");
 for(i=0;i<M;i++)
  {printf(" Number %d suject equation is: %d   %f  ",i+1,Code[i],B[i]);
  for(j=0;j<K;j++)
   {printf("(%f)X%d ",A[i][j],j+1);
   if(j!=K-1)
     printf("+");
   }
  }
 /******   initiate data   ******/
   initiate();
 solve();
 if(!Type)
  A[M][N]=-A[M][N];
 printf(" The optimal value of the original objective function is:   %f  ",A[M][N]);
 getch();
}


 /******   initiate variables function  ******/
void initiate(){
 int i,j;
 Indexg=K;
 Indexl=Indexg+Get;
 Indexe=Indexl+Let;;
 for(i=0;i<M+1;i++)
  for(j=K;j<N+1;j++)
   A[i][j]=0;
 for(i=0;i<M;i++)
  A[i][N]=B[i];
 for(i=0;i<M;i++)
  switch(Code[i])
  { case 0: {XB[i]=Indexl;
     A[i][Indexl++]=1;
     break;
     };
   case 1: {XB[i]=Indexe;
     IAA[IA++]=i;
     A[i][Indexe++]=1;
     A[i][Indexg++]=-1;
     break;
     };
   case 2: {XB[i]=Indexe;
           IAA[IA++]=i;
    A[i][Indexe++]=1;
    break;
    };
  }
 for(j=0;j<K;j++)
  if(Type)
   A[M][j]=-C[j];
  else
   A[M][j]=C[j];
 for(j=K;j<=N;j++)
   A[M][j]=0;
 for(j=K+Get+Let;j<N;j++)
  A[M][j]=Q;
 Sum=0;
 for(j=0;j<=N;j++)
  {Sum=0;
  for(i=0;i<IA;i++)
   Sum=Sum+A[IAA[i]][j];
  A[M][j]=A[M][j]-Sum*Q;
  }
 return;
}

/******   process data  function  ******/
void  solve(){
 int i,j,mark=1,minus,minusmark,basic=0,divide,dividemark;
 float H,P;
 while(1)
  {mark=0;
  minusmark=0;
  minus=0;divide=1000;
  dividemark=0;
  printf(" Basic solution %d is ",++basic);
  for(i=0;i<M;i++)
   printf("Basic variable %d = X( %d )= %f ",i+1,XB[i]+1,A[i][N]);
  printf(" Current value of the object equation is:      %f ",A[M][N]);
  getch();
  for(j=0;j<N;j++)
   {
   if(A[M][j]<-6e-8)
    mark++;

   if(A[M][j]<minus)
    {minus=A[M][j];
    minusmark=j;
    }
    }
  if(mark==0)
         break;
  for(i=0;i<M;i++)
   {if(A[i][minusmark]==0)
    continue;
   if(A[i][N]/A[i][minusmark]<=0)
    continue;
   if(A[i][N]/A[i][minusmark]<divide)
    {divide=A[i][N]/A[i][minusmark];
    dividemark=i;
    }
    }
  XB[dividemark]=minusmark;
  if(divide<0)
   printf("There is no solution because of no boundary!");
                P=A[dividemark][minusmark];
  for(j=0;j<N+1;j++)
   A[dividemark][j]=A[dividemark][j]/P;
  for(i=0;i<M+1;i++)
    {H=A[i][minusmark];
    if(i==dividemark)
     continue;
    for(j=0;j<N+1;j++)
     A[i][j]=A[i][j]-H*A[dividemark][j];
    }
  }
  printf(" ************************************* The last basic solution is optimal! *************************************  ");