All Pair Shortest Path (APSP)

Question : Find shortest paths between all pairs of vertices in a graph.

Floyd-Warshall Algorithm
It is one of the easiest algorithms, and just involves simple dynamic programming. The algorithm can be read from this wikipedia page.

#define SIZE 31
#define INF 1e8
double dis[SIZE][SIZE];
void init(int N)
{
	for (k=0;k<N;k++)
		for (i=0;i<N;i++)
			dis[i][j]=INF;
}
void floyd_warshall(int N)
{
        int i,j,k;
        for (k=0;k<N;k++)
                for (i=0;i<N;i++)
                        for (j=0;j<N;j++)
                                dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);
}

int main()
{
	//input size N
	init(N);
	//set values for dis[i][j]
	floyd_warshall(N);
}

We can also use the algorithm to

1. find the shortest path
   • we can use another matrix called predecessor matrix to construct the shortest path.
2. find negative cycles in a graph.
   • If the value of any of the diagonal elements is less than zero after calling the floyd-warshall algorithm then there is a negative cycle in the graph.
3. find transitive closure
   • to find if there is a path between two vertices we can use a boolean matrix and use and-& and or-| operators in the floyd_warshall algorithm.
   • to find the number of paths between any two vertices we can use a similar algorithm.

NJOY!!
-fR0DDY

Convert a number from decimal base to any Base

Convert a given decmal number to any other base (either positive or negative).

For example, 100 in Base 2 is 1100100 and in Base -2 is 110100100.

Here’s a simple algorithm to convert numbers from Decimal to Negative Bases :

def tonegativeBase(N,B):
   digits = []
   while i != 0:
	i, remainder = divmod (i, B)
	if (remainder < 0):
	    i, remainder = i + 1, remainder + B*-1
	digits.insert (0, str (remainder))
   return ''.join (digits)

We can just tweak the above algorithm a bit to convert a decimal to any Base. Here’s a sample code :

#include<iostream>
using namespace std;

void convert10tob(int N,int b)
{
     if (N==0)
        return;
     
     int x = N%b;
     N/=b;
     if (x<0)
        N+=1;
        
     convert10tob(N,b);
     printf("%d",x<0?x+(b*-1):x);
     return;
}


int main()
{
    int N,b;
    while (scanf("%d%d",&N,&b)==2)
    {
          if (N!=0)
          {
              convert10tob(N,b);
              printf("\n");
          }
          else
              printf("0\n");
    }
}

NJOY!
-fR0DDY