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Pastes found: 2206
12 Jan 2017, 04:37 PM

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#include<bits/stdc++.h>
 
int max(int a, int b);
 
int lcs( char *X, char *Y, int m, int n )
{
   if (m == 0 || n == 0)
     return 0;
   if (X[m-1] == Y[n-1])
     return 1 + lcs(X, Y, m-1, n-1);
   else
     return max(lcs(X, Y, m, n-1), lcs(X, Y, m-1, n));
}
 
int max(int a, int b)
{
    return (a > b)? a : b;
}
 
int main()
{
  char X[] = "AGGTAB";
  char Y[] = "GXTXAYB";
 
  int m = strlen(X);
  int n = strlen(Y);
 
  printf("Length of LCS is %d\n", lcs( X, Y, m, n ) );
 
  return 0;
}
 
 

This is the pop-up flow.

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SOLVE 5
/* A Naive recursive implementation of LCS problem */
#include<bits/stdc++.h>
 
int max(int a, int b);
 
/* Returns length of LCS for X[0..m-1], Y[0..n-1] */
int lcs( char *X, char *Y, int m, int n )
{
   if (m == 0 || n == 0)
     return 0;
   if (X[m-1] == Y[n-1])
     return 1 + lcs(X, Y,

This is the pop-up flow.

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SOLVE 5
/* A Naive recursive implementation of LCS problem */
#include<bits/stdc++.h>
 
int max(int a, int b);
 
/* Returns length of LCS for X[0..m-1], Y[0..n-1] */
int lcs( char *X, char *Y, int m, int n )
{
   if (m == 0 || n == 0)
     return 0;
   if (X[m-1] == Y[n-1])
     return 1 + lcs(X, Y,

This is the pop-up flow.

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solve 4 
#include<stdio.h>
#include<limits.h>
 
// Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
int MatrixChainOrder(int p[], int i, int j)
{
    if(i == j)
        return 0;
    int k;
    int min = INT_MAX;
    int count;
 
    // place parenthesis at different places between first
    // an

This is the pop-up flow.

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SOLVE 6
 * C++ Program to Find All Pairs Shortest Path
 
 */
 
#include <iostream>
 
#include <cstdlib>
 
#define max 10
 
#define infi 999
 
using namespace std;
 
int p[max][max];
 
/*
 
 * All Pairs Shortest Path using Floyd's Algorithm
 
 */
 
void allpairshort(int a[max][max], int n)
 
{
 
    int k, i, j;
 

This is the pop-up flow.

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For ODD:
Solve 2:
#include<iostream>
#include<conio.h>
#include<stdlib.h>
using namespace std;
int cost[10][10],i,j,k,n,qu[10],front,rare,v,visit[10],visited[10];
 
main()
{
int m;
cout <<"enterno of vertices";
cin >> n;
cout <<"ente no of edges";
cin >> m;
cout <<"\nEDGES \n";
for(k=1;k<=m;k++)
{
c

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For even:
Solve 2:
#include<iostream>
#include<conio.h>
#include<stdlib.h>
using namespace std;
int cost[10][10],i,j,k,n,qu[10],front,rare,v,visit[10],visited[10];
 
main()
{
int m;
cout <<"enterno of vertices";
cin >> n;
cout <<"ente no of edges";
cin >> m;
cout <<"\nEDGES \n";
for(k=1;k<=m;k++)
{
c

This is the pop-up flow.

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For Even: 
Solve 1:
#include<stdio.h>
#include<conio.h>
void main()
{
int path[5][5],i,j,min,a[5][5],p,st=1,ed=5,stp,edp,t[5],index;
printf("enter the cost matrix\n");
for(i=1;i<=5;i++)
for(j=1;j<=5;j++)
scanf("%d",&a[i][j]);
printf("enter  number of paths\n");
scanf("%d",&p);
printf("enter possible

This is the pop-up flow.

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#include<stdio.h>
#include<conio.h>
void main()
{
int path[5][5],i,j,min,a[5][5],p,st=1,ed=5,stp,edp,t[5],index;
printf("enter the cost matrix\n");
for(i=1;i<=5;i++)
for(j=1;j<=5;j++)
scanf("%d",&a[i][j]);
printf("enter  number of paths\n");
scanf("%d",&p);
printf("enter possible paths\n");
for(i=1;
08 Jan 2017, 01:34 PM

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