C++ 實習測試: 1042 Quiz#1 互斥集合 (Disjoint Sets) |
令集合 S 為整數 0 到 n-1 的集合,「互斥集合 (disjoint-sets)」資料結構 (或稱為「不相交集合」, 也稱為 union-find 資料結構, 或是 merge-find 資料結構) 的定義如下:
|
我們可以用 陣列 簡單地實作 「互斥集合」資料結構, 用樹狀架構來代表各個子集合, 例如 DS2 = {0, 1, 2, 5}, {3, 6, 7}, {4} 表示為下圖之三個樹狀架構 用樹根節點作為這個子集合的代表元素, 下圖的整數陣列中每一個元素 parent[i] 中紀錄的是節點 i 的父節點, 如果是根節點的話 parent[i] 記錄的是 i 自己, 例如 DS2 = {0, 1, 2, 5}, {3, 6, 7}, {4} 以下列整數陣列表示 注意: 用這種表示方法來表示子集合並不是唯一的,同一個子集合可以有不同的樹狀架構,不考慮效率的情況下這個實作可以滿足下列 main() 函式中的基本測試 |
下面是一個實作 互斥集合 (Disjoint Sets) 的 C 程式 |
#include <cstdlib> #include <cassert> #include <iostream> #include <iomanip> using namespace std; void makeSets(int nNodes, int parent[]); int findSets(int i, int nNodes, const int parent[]); void mergeTrees(int i, int j, int nNodes, int parent[]); void unionSets(int i, int j, int nNodes, int parent[]); int equalSets(int nNodes1, const int parent1[], int nNodes2, const int parent2[]); void printSets(int nNodes, const int parent[]); int main() { const int nNodes=8; int parent[nNodes]; int i; const int p1[8] = {0,1,2,3,4,5,6,7}; const int p2[8] = {0,0,2,3,4,5,6,7}; const int p3[8] = {0,0,3,3,4,5,6,7}; const int p4[8] = {0,0,3,3,4,4,6,7}; const int p5[8] = {0,0,3,3,5,5,5,7}; const int p6[8] = {0,0,3,3,6,4,6,6}; const int p7[8] = {0,0,3,4,6,4,6,6}; const int p8[8] = {7,0,3,4,6,4,6,6}; makeSets(nNodes, parent); assert(equalSets(nNodes, parent, nNodes, p1)); unionSets(0, 1, nNodes, parent); assert(equalSets(nNodes, parent, nNodes, p2)); unionSets(2, 3, nNodes, parent); assert(equalSets(nNodes, parent, nNodes, p3)); unionSets(4, 5, nNodes, parent); assert(equalSets(nNodes, parent, nNodes, p4)); unionSets(4, 6, nNodes, parent); assert(equalSets(nNodes, parent, nNodes, p5)); unionSets(4, 7, nNodes, parent); assert(equalSets(nNodes, parent, nNodes, p6)); unionSets(3, 7, nNodes, parent); assert(equalSets(nNodes, parent, nNodes, p7)); unionSets(2, 1, nNodes, parent); assert(equalSets(nNodes, parent, nNodes, p8)); printSets(nNodes, parent); system("pause"); return 0; } void printSets(int nNodes, const int parent[]) { int i, j; int *printed = (int *) malloc(nNodes*sizeof(int)); for (i=0; i<nNodes; i++) printed[i] = 0; for (i=0; i<nNodes; i++) if (!printed[i]) { if (i==0) cout << "{"; else cout << ", {"; cout << i; for (j=i+1; j<nNodes; j++) if (!printed[j] && findSets(i, nNodes, parent)==findSets(j, nNodes, parent)) { cout << ", " << j; printed[j] = 1; } cout << "}"; } cout << endl; free(printed); } void makeSets(int nNodes, int parent[]) { int i; for (i=0; i<nNodes; i++) parent[i] = i; } int findSets(int i, int nNodes, const int parent[]) { if ((i<0)||(i>=nNodes)) return -1; if (i == parent[i]) return i; return findSets(parent[i], nNodes, parent); } void mergeTrees(int i, int j, int nNodes, int parent[]) { if ((i<0)||(i>=nNodes)) return; if ((j<0)||(j>=nNodes)) return; parent[i] = j; } void unionSets(int i, int j, int nNodes, int parent[]) { if ((i<0)||(i>=nNodes)) return; if ((j<0)||(j>=nNodes)) return; mergeTrees(findSets(i, nNodes, parent), findSets(j, nNodes, parent), nNodes, parent); } int equalSets(int nNodes1, const int parent1[], int nNodes2, const int parent2[]) { if (nNodes1 != nNodes2) return 0; for (int i=0; i<nNodes1; i++) { if (findSets(findSets(i, nNodes1, parent1), nNodes1, parent2) != findSets(i, nNodes1, parent2)) return 0; if (findSets(findSets(i, nNodes1, parent2), nNodes1, parent1) != findSets(i, nNodes1, parent1)) return 0; } return 1; } |
下面是一個實作 互斥集合 (Disjoint Sets) 的 C++ 物件化程式 |
/* DisjointSets.h */ #pragma once class DisjointSets { public: DisjointSets(void); virtual ~DisjointSets(void); void makeSets(int nNodes, const int *parent=0); int findSets(int i) const; void unionSets(int i, int j); int equalSets(const DisjointSets& rhs) const; void print(); static void unitTest(void); private: void mergeTrees(int i, int j); int m_nNodes; int m_parent[100]; }; /* DisjointSets.cpp */ #include "DisjointSets.h" #include <cassert> #include <iostream> using namespace std; DisjointSets::DisjointSets(void) { } DisjointSets::~DisjointSets(void) { } void DisjointSets::makeSets(int nNodes, const int *parent) { int i; m_nNodes = nNodes; if (parent == 0) for (i=0; i<nNodes; i++) m_parent[i] = i; else for (i=0; i<nNodes; i++) m_parent[i] = parent[i]; } int DisjointSets::findSets(int i) const { if ((i<0)||(i>=m_nNodes)) return -1; if (i == m_parent[i]) return i; return findSets(m_parent[i]); } void DisjointSets::mergeTrees(int i, int j) { if ((i<0)||(i>=m_nNodes)) return; if ((j<0)||(j>=m_nNodes)) return; m_parent[i] = j; } void DisjointSets::unionSets(int i, int j) { if ((i<0)||(i>=m_nNodes)) return; if ((j<0)||(j>=m_nNodes)) return; mergeTrees(findSets(i), findSets(j)); } int DisjointSets::equalSets(const DisjointSets& rhs) const { int i; if (m_nNodes != rhs.m_nNodes) return 0; for (i=0; i<m_nNodes; i++) if ((rhs.findSets(findSets(i)) != rhs.findSets(i)) || (findSets(rhs.findSets(i)) != findSets(i))) return 0; return 1; } void DisjointSets::unitTest(void) { const int nNodes=8; int i; const int p1[8] = {0,1,2,3,4,5,6,7}; const int p2[8] = {0,0,2,3,4,5,6,7}; const int p3[8] = {0,0,3,3,4,5,6,7}; const int p4[8] = {0,0,3,3,4,4,6,7}; const int p5[8] = {0,0,3,3,5,5,5,7}; const int p6[8] = {0,0,3,3,6,4,6,6}; const int p7[8] = {0,0,3,4,6,4,6,6}; const int p8[8] = {7,0,3,4,6,4,6,6}; DisjointSets ds1, ds2; ds1.makeSets(nNodes); ds2.makeSets(nNodes, p1); assert(ds1.equalSets(ds2)); ds1.unionSets(0, 1); ds2.makeSets(nNodes, p2); assert(ds1.equalSets(ds2)); ds1.unionSets(2, 3); ds2.makeSets(nNodes, p3); assert(ds1.equalSets(ds2)); ds1.unionSets(4, 5); ds2.makeSets(nNodes, p4); assert(ds1.equalSets(ds2)); ds1.unionSets(4, 6); ds2.makeSets(nNodes, p5); assert(ds1.equalSets(ds2)); ds1.unionSets(4, 7); ds2.makeSets(nNodes, p6); assert(ds1.equalSets(ds2)); ds1.unionSets(3, 7); ds2.makeSets(nNodes, p7); assert(ds1.equalSets(ds2)); ds1.unionSets(2, 1); ds2.makeSets(nNodes, p8); assert(ds1.equalSets(ds2)); ds1.print(); } void DisjointSets::print() { int i, j, printed[100]; for (i=0; i<m_nNodes; i++) printed[i] = 0; for (i=0; i<m_nNodes; i++) if (!printed[i]) { if (i==0) cout << "{"; else cout << ", {"; cout << i; for (j=i+1; j<m_nNodes; j++) if (!printed[j] && findSets(i)==findSets(j)) { cout << ", " << j; printed[j] = 1; } cout << "}"; } cout << endl; } /* main.cpp */ #include "DisjointSets.h" #include <cstdlib> int main() { DisjointSets::unitTest(); system("pause"); return 0; } |
回
C++ 物件導向程式設計課程
首頁
製作日期: 03/25/2016 by 丁培毅 (Pei-yih Ting) E-mail: [email protected] TEL: 02 24622192x6615 海洋大學 電機資訊學院 資訊工程學系 Lagoon |