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我想爲我的代碼的單元測試隨機生成樹(而不是BST)。我嘗試了很多方法,但是在生成3到4棵樹後出現異常,或者代碼進入了無限循環。我正在使用隨機數的邊和節點值。
我也嘗試過使用隨機數字填充隊列的隊列方法,然後使隊列中的隊列中的現有項目隨機選擇舊節點,然後連接此新節點。爲單元測試隨機生成樹
有人知道如何以更好更簡潔的方式在C#中執行此操作嗎?
編輯
public class Tree
{
public Node Root { get; private set; }
public readonly Dictionary<long, Node> Nodes = new Dictionary<long, Node>();
public readonly Dictionary<string, long> SumDictionary = new Dictionary<string, long>();
public readonly Dictionary<long, long> FDictionary = new Dictionary<long, long>
{
{0, 1},
{1, 1}
};
public long Query(long n)
{
long s = 0;
Dfs();
for (int i = 1; i <= n; i++)
{
for (int j = i; j <= n; j++)
{
var lvalue = GetLValue(i, j);
if (i != j)
{
lvalue *= 2;
}
s += lvalue;
if (s >= Mod)
{
s %= Mod;
}
}
}
return s;
}
private long GetLValue(int a, int b)
{
var key = string.Format("{0}-{1}", a, b);
if (SumDictionary.ContainsKey(key))
{
return Fvalue(SumDictionary[key]);
}
var aKey = string.Format("{0}-{1}", 1, a);
var bKey = string.Format("{0}-{1}", 1, b);
var sumA = SumDictionary[aKey];
var sumB = SumDictionary[bKey];
var lca = Lca(a, b);
if (lca != a && lca != b)
{
return Fvalue(sumA + sumB - SumDictionary[string.Format("{0}-{1}", lca, lca)]);
}
long bigSum, smallSum;
if (sumA > sumB)
{
bigSum = sumA;
smallSum = sumB;
}
else
{
bigSum = sumB;
smallSum = sumA;
}
var sumAtoB = bigSum - (smallSum - Nodes[smallSum].Value);
return Fvalue(sumAtoB);
}
public void AddNode(long a, long b)
{
Node nodea, nodeb;
if (Nodes.ContainsKey(a))
{
nodea = Nodes[a];
}
else
{
nodea = new Node { Value = a };
Nodes.Add(a, nodea);
}
if (Nodes.ContainsKey(b))
{
nodeb = Nodes[b];
}
else
{
nodeb = new Node { Value = b };
Nodes.Add(b, nodeb);
}
nodea.ReachableNodes.Add(b, nodeb);
nodeb.ReachableNodes.Add(a, nodea);
if (Root == null)
{
Root = nodea;
}
}
public long? Lca(int a, int b)
{
bool found = false;
return TraverseForLca(Root, null, a, b, ref found);
}
private long? TraverseForLca(Node node, Node prev, long a, long b, ref bool found)
{
if (node == null)
{
return null;
}
if (node.Value == a)
{
return a;
}
if (node.Value == b)
{
return b;
}
long? f = null;
foreach (KeyValuePair<long, Node> reachableNode in node.ReachableNodes)
{
var n = reachableNode.Value;
if (prev != null && n.Value == prev.Value)
{
continue;
}
long? lca = TraverseForLca(n, node, a, b, ref found);
if (found)
{
return lca;
}
if (lca != null && f == null)
{
f = lca;
}
else if (lca != null)
{
found = true;
return node.Value;
}
}
return f;
}
public void Dfs()
{
TravelForDfs(Root, null, 0);
}
private void TravelForDfs(Node node, Node prev, long recSum)
{
if (node == null)
{
return;
}
var key = string.Format("{0}-{1}", prev != null ? prev.Value : node.Value, node.Value);
var iKey = string.Format("{0}-{1}", node.Value, node.Value);
var weight = node.Weight;
if (weight >= Mod)
{
weight %= Mod;
}
if (!SumDictionary.ContainsKey(iKey))
{
SumDictionary.Add(iKey, weight);
}
weight = recSum + node.Weight;
if (weight >= Mod)
{
weight %= Mod;
}
foreach (KeyValuePair<long, Node> reachableNode in node.ReachableNodes)
{
var n = reachableNode.Value;
if (prev != null && n.Value == prev.Value)
{
if (!SumDictionary.ContainsKey(key))
{
SumDictionary.Add(key, weight);
}
continue;
}
if (!SumDictionary.ContainsKey(key))
{
SumDictionary.Add(key, weight);
}
TravelForDfs(n, node, weight);
}
}
public long Fvalue(long n)
{
if (n == 1 || n == 0)
{
return 1;
}
long a, b;
if (FDictionary.ContainsKey(n - 1))
{
a = FDictionary[n - 1];
}
else
{
a = Fvalue(n - 1);
if (!FDictionary.ContainsKey(n - 1))
{
FDictionary.Add(n - 1, a);
}
}
if (FDictionary.ContainsKey(n - 2))
{
b = FDictionary[n - 2];
}
else
{
b = Fvalue(n - 2);
if (!FDictionary.ContainsKey(n - 2))
{
FDictionary.Add(n - 2, b);
}
}
if (!FDictionary.ContainsKey(n))
{
FDictionary.Add(n, a + b);
}
var s = a + b;
if (s >= Mod)
{
s %= Mod;
}
return s;
}
}
請郵寄我不想隨機樹生成功能依賴於問題的樹的代碼和你想要什麼 – opewix
一個例子。它應該只生成一棵樹。就像一串隨機數字一樣。完全獨立於問題。無論如何,我已經發布了我的樹代碼。 –
請解釋'AddNode'方法的邏輯,爲什麼它接受兩個參數,爲什麼你需要'Dictionary' –
opewix