static long roadsAndLibraries(int n, long c_lib, long c_road, int[][] cities)
{
GraphAPI gapi = new GraphAPI(n);
for (int i = 0; i < cities.Length; i++)
{
gapi.AddEdge(cities[i][0], cities[i][1]);
}
ConnectedComponents bfs = new ConnectedComponents(gapi, n);
long result = ((c_road * bfs.RoadCount()) + (bfs.LibraryCount() * c_lib));
if (c_lib * n < result)
{
result = c_lib * n;
}
return(result);
}
/*Suppose that we have n groups, of sizes a1 to an. Except when n is smallish, the following formula for the number N of ways
* to choose a pair of people from different groups is more efficient:
* N=((a1+⋯+an)^2 − (a1^2 +⋯+ an^2)) / 2.
* To see that the formula does the job, expand (a1+⋯+an)2.*/
static long journeyToMoon(int n, int[][] astronaut)
{
GraphAPI gapi = new GraphAPI(n);
for (int i = 0; i < astronaut.Length; i++)
{
gapi.AddEdge(astronaut[i][0], astronaut[i][1]);
}
ConnectedComponents bfs = new ConnectedComponents(gapi, n);
IList <long> IdCount = new List <long>();
IdCount.Add(1);
Array.Sort(bfs.Id);
for (int k = 0; k < n - 1; k++)
{
if (bfs.Id[k] == bfs.Id[k + 1])
{
++IdCount[IdCount.Count - 1];
continue;
}
IdCount.Add(1);
}
long SumSquare = 0;
for (int i = 0; i < IdCount.Count; i++)
{
SumSquare += IdCount[i];
}
SumSquare = (long)Math.Pow(SumSquare, 2);
long SquareSum = 0;
for (int i = 0; i < IdCount.Count; i++)
{
SquareSum += (long)Math.Pow(IdCount[i], 2);
}
return((SumSquare - SquareSum) / 2);
}
