anshumanatri

#include<iostream.h>
#include<conio.h>
#include<stdio.h>
#include<stdlib.h>
int h[10],n;
void heap(int m);
void main()
{
 clrscr();
 int i,m;

anshumanatri

#include<iostream.h>
#include<conio.h>
#include<stdio.h>
int w[10][10],n,ner[10],visit[10],dist[10],source;
void dijkstra()
{
  for(int i=1;i<n-1;i++)
  {
   int min=1234,u=0;
   for(int j=1;j<=n;j++)

anshumanatri

#include<iostream.h>
#include<conio.h>
#include<stdio.h>
int b[10][10];
int binomial(int n, int k)
{
  int i,j;
  for(i=0;i<=n;i++)
   for(j=0;j<=(i<k?i:k);j++)
    if(j==0 || j==i)

anshumanatri

#include<iostream.h>
#include<stdio.h>
#include<conio.h>
int q[10],e[10][10],n,order[10],count=1,front=0,rear=-1;
void bfs(int node)
{
   int i;
   q[++rear]=node;
   order[node]=count;
   while(count!=n)

anshumanatri

//Tower of Hanoi
//No of calls made for n disks is equal to pow(2,n) -1.

   void  hanoi(int n, char a, char c, char b)
     {
       if(n==1)
       {
	printf("\nShift disk 1 from %c to %c",a,c);
	return  ;
	}

anshumanatri

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
#define null 0
struct btree
{
int data;
struct btree * left;
struct btree * right;
};

anshumanatri

#include<stdio.h>
#include<conio.h>
#include<ctype.h>
void main()
{  static int a[5];
   int i,n,top=-1,op1,op2,noperator=0,noperand=0;
   char ch[10];
   clrscr();
   printf("\n Enter the string\n");
   gets(ch);

anshumanatri

#include<stdio.h>
#include<conio.h>
#include<ctype.h>
void main()
{
  int i,j=-1,top=-1;
  clrscr();
  char in[30], po[30],stack[20];
  printf("\n Enter the infix expression ");
  gets(in);

anshumanatri

void quicksort(int l, int u)
{   int i, m;
    if (l >= u) return;
    swap(l, randint(l, u));
    m = l;
    for (i = l+1; i <= u; i++)
        if (x[i] < x[l])
            swap(++m, i);
    swap(l, m);
    quicksort(l, m-1);

anshumanatri

Change(d,k,n)
c[0]  = 0
for p = 1 to n do
    min  =  INFINITY
    for  i =  1 to  k    do
         if(d[i] <= p) do 
            if( 1 + c[p - d[i]] < min) then
                min = 1 + c[p - d[i]]
                coin = i
     c[p] = min