public void TestMethod2()
{
string[] terrain = new string[]
{
"P...S"
, "..#.."
, "....."
, "....."
, "....D"
};
BraveKnight braveKnight = new BraveKnight(terrain);
Assert.IsTrue(braveKnight.Princess.Row == 0 && braveKnight.Princess.Column == 0);
Assert.IsTrue(braveKnight.Start.Row == 0 && braveKnight.Start.Column == 4);
Assert.IsTrue(braveKnight.Destination.Row == 4 && braveKnight.Destination.Column == 4);
Assert.IsTrue(braveKnight.graph[0].Keys.Count == 1);
Assert.IsTrue(braveKnight.graph[0].ContainsKey(11));
Assert.IsTrue(braveKnight.graph[4].Keys.Count == 1);
Assert.IsTrue(braveKnight.graph[4].ContainsKey(13));
Assert.IsTrue(braveKnight.graph[12].Keys.Count == 8);
Assert.IsTrue(braveKnight.graph[12].ContainsKey(9));
Assert.IsTrue(braveKnight.graph[12].ContainsKey(19));
}
public void TestMethod5()
{
string[] terrain = new string[]
{
"P...S"
, "..#.."
, ".#..."
, "....."
, "....D"
};
BraveKnight braveKnight = new BraveKnight(terrain);
Assert.IsFalse(braveKnight.MinDistance(0, 24).HasValue);
}
public void TestMethod4()
{
string[] terrain = new string[]
{
"P...S"
, "..#.."
, "....."
, "....."
, "....D"
};
BraveKnight braveKnight = new BraveKnight(terrain);
Assert.IsTrue(braveKnight.MinDistance(0, 4) == 4);
}
public void TestMethod8()
{
string[] terrain = new string[]
{
"P...S"
, "....."
, "....."
, "....."
, "..*.D"
};
BraveKnight braveKnight = new BraveKnight(terrain);
Assert.IsTrue(braveKnight.MinDistance(24, 0) == 3);
}
public void TestMethod6()
{
string[] terrain = new string[]
{
"P...S"
, "....."
, "....."
, "....."
, "....D"
};
BraveKnight braveKnight = new BraveKnight(terrain);
braveKnight.ExistShortestRescue(out int distance);
Assert.IsTrue(distance == 6);
}
/// <summary>
/// Explanation :
/// the BraveKnight class is the Main Class .
/// The Idea is to make a Directed Graph with all the squares .
/// Each square is a vertex . There would be an edge (v1,v2) if the vertex v1 could go to vertex v2 in just one step.
/// Each edge has the same value, then we could use a Breadth-first search and Shortest-Distance algorithm
/// when we want to calculate the shortest-path in steps from a vertex v to w .(Maybe it doesn't exist)
/// The solution is calculating first the shortest-path from Start Position -> to Princess Position (if Exists)
/// then From Princess Position to Destination Position ( if exists) the firstDistance + secondDistance is the answer.
/// In other case is IMPOSSIBLE.
/// graph G = (V = vertexs = N*M ,E = edges = O(V) )
/// Each vertex has at most 8 edges => E = O(V)
/// this algorithm is O(V+E) = O(V) = O (N*M) because E = O(V) and BFS = O(V+E)
/// The result is O(C*N*M)
/// </summary>
/// <param name="args"></param>
public static void Main(string[] args)
{
try
{
int C = int.Parse(Console.ReadLine());
for (int i = 1; i <= C; i++)
{
try
{
var line = Console.ReadLine();
string[] splitParts = line.Split(new char[] { ' ' }, StringSplitOptions.RemoveEmptyEntries);
int N = int.Parse(splitParts[0]);
int M = int.Parse(splitParts[1]);
string[] terrain = new string[N];
for (int j = 0; j < N; j++)
{
line = Console.ReadLine();
terrain[j] = line;
}
BraveKnight braveKnight = new BraveKnight(terrain);
string prefixString = "Case #" + i + ": ";
if (braveKnight.ExistShortestRescue(out int result))
{
Console.WriteLine(prefixString + result);
}
else
{
Console.WriteLine(prefixString + "IMPOSSIBLE");
}
}
catch (Exception exception)
{
Console.WriteLine("Case #" + i + ": ");
Console.WriteLine(exception);
}
}
}
catch (Exception exception)
{
Console.WriteLine(exception);
}
}
