有向图(7)--再谈可达性&&有向图总结

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顶点对的可达性。给定一幅有向图，回答“是否存在一条从一个给定的顶点v到另一个给定的顶点w的路径？”等类似问题。

可以使用深度搜索来实现，无论对于稀疏还是稠密的图，它都是理想的解决方案，但它不适用于在实际应用中可能遇到的大型有向图，因为所需的空间和V²成正比，所需时间和V(V+E)成正比。

-TransitiveClosure.h

#ifndef __TRANSITIVE_CLOSURE_H__#define __TRANSITIVE_CLOSURE_H__#include "Digraph.h"#include "DirectedDFS.h"class TransitiveClosure {private:DirectedDFS* all;public:TransitiveClosure(Digraph G);<span style="white-space:pre"></span>~TransitiveClosure() { delete[] all; }bool reachable(int v, int w) { return all[v].isMarked(w); }};TransitiveClosure::TransitiveClosure(Digraph G) {all = new DirectedDFS[G.getV()];for (int v = 0; v < G.getV(); ++v)all[v].createDFS(G, v);}#endif

这里我稍微对之前的DirectedDFS做了修改，还有给Digraph的构造函数添加了默认值，因为new的时候只能调用默认构造函数，所以写了新的函数来重新构造bool* marked

-DirectedDFS.h

#ifndef __DIRECTED_DFS_H__#define __DIRECTED_DFS_H__#include "Digraph.h"// 解决单点可达性或者多点可达性class DirectedDFS {private:bool* marked;public:DirectedDFS(const Digraph& G = Digraph(), int s = 0);DirectedDFS(const Digraph& G, std::list<int> ilst);~DirectedDFS() { delete[] marked; }bool isMarked(int v)const { return marked[v]; }void createDFS(Digraph G, int s);// 为了在TransitiveClosure中重新构造private:void dfs(const Digraph& G, int v);};// constructorDirectedDFS::DirectedDFS(const Digraph& G, int s) {if (0 == G.getV()) return;marked = new bool[G.getV()];for (int i = 0; i < G.getV(); ++i)marked[i] = false;dfs(G, s);}// constructorDirectedDFS::DirectedDFS(const Digraph& G, std::list<int> ilst) {marked = new bool[G.getV()];for (int i = 0; i < G.getV(); ++i)marked[i] = false;for (int v : ilst) {if (!marked[v])dfs(G, v);}}void DirectedDFS::dfs(const Digraph& G, int v) {marked[v] = true;for (int w : G.getAdj(v)) {if (!marked[w])dfs(G, w);}}// 重新构造DirectedDFS，和构造函数差不多了void DirectedDFS::createDFS(Digraph G, int s) {if(NULL == marked) delete[] marked;marked = new bool[G.getV()];for (int i = 0; i < G.getV(); ++i)marked[i] = false;dfs(G, s);}#endif

测试用例 -main.cpp

#include "TransitiveClosure.h"#include "Digraph.h"#include <iostream>using namespace std;int main(){int vNum, eNum;cin >> vNum >> eNum;Digraph G(vNum);int v, w;for (int i = 0; i < eNum; ++i) {cin >> v >> w;G.addEdge(v, w);}TransitiveClosure tc(G);int a, b;for (;;) {cout << "输入起点：";cin >> a;cout << "输入终点：";cin >> b;cout << "                      " << a << " 到 " << b << "是";if (!tc.reachable(a, b)) cout << "不";cout << "可达的" << endl;}return 0;}

有向图总结

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