技術標籤：日常練習dfs連通子圖最短路徑

最短路徑

從指定起點s做廣度優先搜尋，總能找到一條從s到v的最短路徑，O(V+E)。
標頭檔案 Paths.h

#include"Paths.h"
void findShortestPath(Graph& G, int s, int v) {
	Paths pathG(G, s);//預設廣搜，下面列印最短路徑
	if (pathG.hasPathto(v)) {
		for (auto& w : pathG.pathTo(v)) {
			if (w == s) printf_s("%2d "
 
, s);
			else
				printf_s("-%2d ", w);
		}
		printf_s("\n");
	}
	else {
		printf_s("s v unconnected.\n");
	}
}

//test
int main() {
	Graph G("data.txt");
	findShortestPath(G, 11, 4);
	system("pause");
	return 0;
}

連通子圖的個數

對每個未標記的節點做深度優先搜尋，搜尋次數即連通子圖個數，O(V+E)。

#include"Graph.h"
void dfs(Graph&, int, vector<bool>*);
int numOfConnectedSubgraphs(Graph &G) {
	int ans = 0;
	vector<bool>* marked = new vector<bool>(G.V(), false);
	for (int i = 0; i < G.V(); ++i) {
		if (!marked->at(i)) {
			dfs(G, i,marked);
			++
 
ans;
		}
	}
	return ans;
}

void dfs(Graph& G, int v, vector<bool>* marked) {
	marked->at(v) = true;
	for (auto& next : G.adj(v)) {
		if(!marked->at(next))
			dfs(G, next,marked);
	}
}

int main() {
	Graph G("data.txt");
	cout << numOfConnectedSubgraphs(G) << endl;
	system("pause");
	return 0;
}

連通分量的API 及其實現

標頭檔案見 Graph.h

cc.h

#pragma once
#include"Graph.h"

#define out(x) cout<<x<<" "
#define hh printf_s("\n")
//聯通分量
class CC
{
public:
	CC(Graph& G);

	bool connected(int v, int w) { return m_id->at(v) == (*m_id)[w]; }
	int count() { return m_count; }
	int id(int v) { return m_id->at(v); }
private:
	vector<int>* m_visited = nullptr;
	vector<int>* m_id = nullptr;//屬於哪一個聯通分量： 0 ~ m_count-1
	int m_count = 0;

	void dfs(Graph& G, int start);
};

void testCC();

cc.cpp

#include "CC.h"

CC::CC(Graph& G)
{
	int n = G.V();
	m_visited = new vector<int>(n, 0);
	m_id = new vector<int>(n, -1);

	for (int s = 0; s < n; ++s) {
		if (!m_visited->at(s)) {
			dfs(G, s);
			m_count++;
		}
	}
}

void CC::dfs(Graph& G, int start)
{
	m_visited->at(start) = 1;
	m_id->at(start) = m_count;//當前深度搜索中發現的都是同一個id

	for (auto& next : G.adj(start))
	{
		if (!m_visited->at(next)) {
			dfs(G, next);
		}
	}
}

void testCC()
{
	Graph G("data.txt");
	CC search(G);

	out("0 10 connected ? ");
	out((search.connected(0, 10) ? "yes" : "no")),hh;
	out("0  belongs to subG_i "), out(search.id(0)), hh;
	out("10 belongs to subG_i "), out(search.id(10)), hh;
	out("How many ConnectedSubgraphs ? "), out(search.count()), hh;
}