#include<iostream>
#include<cassert>
#include<ctime>
using namespace std;
class UnionFind
{
private:
int* parent;
int* rank;//rank[i]表示以i为根节点的集合所表示的数的层数
int count;
public:
UnionFind(int count)
{
this->count = count;
parent = new int[count];
rank = new int[count];
for (int i = 0; i < count; i++)
{
parent[i] = i;
rank[i] = 1;//初始的时候每个集合互相独立,假设每一个初始的节点在第一层层数从一开始
}
}
~UnionFind()
{
delete[] parent;
delete[] rank;
}
int find(int p)
{
assert(p >= 0 && p < count);
if (p != parent[p])
{
parent[p] = find(parent[p]);
}
return parent[p];//这里暗含了根节点的父亲就是自己
}
bool isConnected(int p, int q)
{
return find(p) == find(q);
}
void UnionElements(int p, int q)
{
int pRoot = find(p);
int qRoot = find(q);
if (pRoot == qRoot)
{
return;
}
if (rank[pRoot] < rank[qRoot])
{
parent[pRoot] = qRoot;
}
else if(rank[qRoot]<rank[pRoot])
{
parent[qRoot] = pRoot;
}
else
{
//rank[pRoot]==rank[qRoot]
parent[pRoot] = qRoot;
rank[qRoot] += 1;//p接到q根节点的后面,q的总体层数加一
}
}
};
void testUF06(int n)
{
srand(time(NULL));
UnionFind uf = UnionFind(n);
for (int i = 0; i < n; i++)
{
int a = rand() % n;
int b = rand() % n;
uf.UnionElements(a, b);
}
for (int i = 0; i < n; i++)
{
int a = rand() % n;
int b = rand() % n;
cout << a <<"\t"<< b <<"\t"<< uf.isConnected(a, b) << endl;;
}
}
int main()
{
testUF06(10000);
system("pause");
return 0;
}
并查集:实现方式⑥(基于递归的路径压缩)
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转载自blog.csdn.net/dosdiosas_/article/details/105557373
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