C++實現並查集
阿新 • • 發佈:2020-07-06
本文例項為大家分享了C++實現並查集的具體程式碼,供大家參考,具體內容如下
#include <iostream> #include <vector> #include <cassert> using namespace std; class UnionFind{ private: vector<int> parent; int count; //優化,記錄p和q所在組的深度,在合併時將深度小的結點的根指向深度大的結點的根 vector<int> rank; public: UnionFind(int count){ parent.resize(count); rank.resize(count); this->count = count; for(int i = 0; i < count; ++i){ parent[i] = i; rank[i] = 1; } } ~UnionFind(){ parent.clear(); rank.clear(); } //路徑壓縮 int find(int p){ assert(p >= 0 && p < count); if(p != parent[p]) parent[p] = find(parent[p]); return parent[p]; } bool isConnected(int p,int q){ return find(p) == find(q); } void unionElement(int p,int q){ int pRoot = find(p),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{ //兩者的rank相等 parent[pRoot] = qRoot; rank[qRoot] += 1; } } };
小編再補充一段程式碼,之前收藏的一段程式碼:
#include <iostream> using namespace std; class UF { //cnt is the number of disjoint sets. //id is an array that records distinct identity of each set,when two sets are merged,their id will be same. //sz is an array that records the child number of each set including the set self. int *id,cnt,*sz; public: // Create an empty union find data structure with N isolated sets. 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; } } ~UF() { delete[] id; delete[] sz; } // Return the id of component corresponding to object p. int find(int p) { if (p != id[p]){ id[p] = find(id[p]); } return id[p]; } // Replace sets containing x and y with their union. void merge(int x,int y) { int i = find(x); int j = find(y); if (i == j) return; // make smaller root point to larger one if (sz[i] < sz[j]) { id[i] = j; sz[j] += sz[i]; } else { id[j] = i; sz[i] += sz[j]; } cnt--; } // Are objects x and y in the same set? bool connected(int x,int y) { return find(x) == find(y); } // Return the number of disjoint sets. int count() { return cnt; } }; void main(){ UF test(5); test.merge(2,3); test.merge(3,4); cout << test.find(4); cout << test.count(); }
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