C++ program to solve the optimal binary search tree problem
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/* Write a C++ program that uses dynamic programming algorithm to solve the optimal binary search tree problem */
#include
#include
#include
using namespace std;
#define MAX 10
int find(int i,int j);
void print(int,int);
int p[MAX],q[MAX],w[10][10],c[10][10],r[10][10],i,j,k,n,m;
char idnt[7][10];
main()
{
cout << "enter the no, of identifiers";
cin >>n;
cout <<"enter identifiers";
for(i=1;i<=n;i++)
gets(idnt[i]);
cout <<"enter success propability for identifiers";
for(i=1;i<=n;i++)
cin >>p[i];
cout << "enter failure propability for identifiers";
for(i=0;i<=n;i++)
cin >> q[i];
for(i=0;i<=n;i++)
{
w[i][i]=q[i];
c[i][i]=r[i][i]=0;
w[i][i+1]=q[i]+q[i+1]+p[i+1];
r[i][i+1]=i+1;
c[i][i+1]=q[i]+q[i+1]+p[i+1];
}
w[n][n]=q[n];
r[n][n]=c[n][n]=0;
for(m=2;m<=n;m++)
{
for(i=0;i<=n-m;i++)
{
j=i+m;
w[i][j]=w[i][j-1]+p[j]+q[j];
k=find(i,j);
r[i][j]=k;
c[i][j]=w[i][j]+c[i][k-1]+c[k][j];
}
}
cout <<"\n";
print(0,n); }
int find(int i,int j)
{
int min=2000,m,l;
for(m=i+1;m<=j;m++)
if(c[i][m-1]+c[m][j]
{
min=c[i][m-1]+c[m][j];
l=m;
}
return l;
}
void print(int i,int j)
{
if(i
puts(idnt[r[i][j]]);
else
return;
print(i,r[i][j]-1);
print(r[i][j],j);
}
OUTPUTenter the no, of identifiers4
enter identifiersdo
if
int
while
enter success propability for identifiers3 3 1 1
enter failure propability for identifiers2 3 1 1 1
tree in preorder form
if
do
int
while
#include
#include
#include
using namespace std;
#define MAX 10
int find(int i,int j);
void print(int,int);
int p[MAX],q[MAX],w[10][10],c[10][10],r[10][10],i,j,k,n,m;
char idnt[7][10];
main()
{
cout << "enter the no, of identifiers";
cin >>n;
cout <<"enter identifiers";
for(i=1;i<=n;i++)
gets(idnt[i]);
cout <<"enter success propability for identifiers";
for(i=1;i<=n;i++)
cin >>p[i];
cout << "enter failure propability for identifiers";
for(i=0;i<=n;i++)
cin >> q[i];
for(i=0;i<=n;i++)
{
w[i][i]=q[i];
c[i][i]=r[i][i]=0;
w[i][i+1]=q[i]+q[i+1]+p[i+1];
r[i][i+1]=i+1;
c[i][i+1]=q[i]+q[i+1]+p[i+1];
}
w[n][n]=q[n];
r[n][n]=c[n][n]=0;
for(m=2;m<=n;m++)
{
for(i=0;i<=n-m;i++)
{
j=i+m;
w[i][j]=w[i][j-1]+p[j]+q[j];
k=find(i,j);
r[i][j]=k;
c[i][j]=w[i][j]+c[i][k-1]+c[k][j];
}
}
cout <<"\n";
print(0,n); }
int find(int i,int j)
{
int min=2000,m,l;
for(m=i+1;m<=j;m++)
if(c[i][m-1]+c[m][j]
{
min=c[i][m-1]+c[m][j];
l=m;
}
return l;
}
void print(int i,int j)
{
if(i
puts(idnt[r[i][j]]);
else
return;
print(i,r[i][j]-1);
print(r[i][j],j);
}
OUTPUTenter the no, of identifiers4
enter identifiersdo
if
int
while
enter success propability for identifiers3 3 1 1
enter failure propability for identifiers2 3 1 1 1
tree in preorder form
if
do
int
while
This post was written by: Rajendra Prasad
Rajendra Prasad is a professional blogger, web designer and front end web developer. Follow him on Facebook
