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Simple Union-Find Datastructure in C#
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public class UF | |
{ | |
public int[] id { get; set; } | |
public int[] sz { get; set; } | |
public int cnt { get; set; } | |
public UF(int N) | |
{ | |
cnt = N; | |
id = new int[N]; | |
sz = new int[N]; | |
for (int i = 0; i < N; i++) | |
{ | |
id[i] = i; | |
sz[i] = 1; | |
} | |
} | |
// Return the id of component corresponding to object p. | |
public int find(int p) | |
{ | |
int root = p; | |
while (root != id[root]) | |
root = id[root]; | |
while (p != root) | |
{ | |
int newp = id[p]; | |
id[p] = root; | |
p = newp; | |
} | |
return root; | |
} | |
public void merge(int x, int y) | |
{ | |
int i = find(x); | |
int j = find(y); | |
if (i == j) return; | |
if (sz[i] < sz[j]) | |
{ | |
id[i] = j; | |
sz[j] += sz[i]; | |
} | |
else | |
{ | |
id[j] = i; | |
sz[i] += sz[j]; | |
} | |
cnt--; | |
} | |
public bool connected(int x, int y) | |
{ | |
return find(x) == find(y); | |
} | |
public int count() | |
{ | |
return cnt; | |
} | |
} |
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