C++实现dijkstra单源最短路径

Posted 2022-01-03 唐火

代码如下:

#include <iostream>
using namespace std;
const int N = 30;
typedef char ElemType;
const double noEdge = 99999;

class Graph

private:
	double G[N][N];
	int vertexN, edgeN;
	double dist[N];
	bool vis[N];
	int path[N];
	int sv;
	ElemType data[N];

	int findMinDist()
	
		double minDist = noEdge;
		int v = -1;
		for (int i = 0; i < vertexN; i++)
		
			if (!vis[i] && dist[i] < minDist)
			
				minDist = dist[i];
				v = i;
			
		
		return v;
	

public :
	Graph()
	
		for (int i = 0; i < N; i++)
		
			for (int j = 0; j < N; j++)
			
				G[i][j] = noEdge;
			
		
		cout << "vertexN = ? and edgeN = ?" << endl;
		cin >> vertexN >> edgeN;
		for (int i = 0; i < vertexN; i++)
		
			cin >> data[i];
		
		cout << "Please input information of edge(including weight)" << endl;
		for (int i = 0; i < edgeN; i++)
		
			int v1, v2;
			double w;
			cin >> v1 >> v2 >> w;
			G[v1][v2] = w;
		
	




	bool dijkstra(int s)
	
		sv = s;
		for (int i = 0; i < vertexN; i++)
		
			dist[i] = G[s][i];
			if (dist[i] < noEdge)
			
				path[i] = s;
			
			else
			
				path[i] = -1;
			
			vis[i] = false;
		
		dist[s] = 0;
		vis[s] = true;

		while (true)
		
			int v = findMinDist();
			if (v == -1) break;

			vis[v] = true;
			for (int w = 0; w < vertexN; w++)
			
				if (!vis[w] && G[v][w] < noEdge)
					if (G[v][w] < 0) return false;
				if (dist[w] > dist[v] + G[v][w])
				
					dist[w] = dist[v] + G[v][w];
					path[w] = v;
				
			
		
		return true;
	


	void printPath(int o)
	

		cout << "value = " << dist[o] << endl;
		if (dist[o] < noEdge)
		
			while (path[o] != sv)
			
				cout << data[o] << " <- " << data[path[o]] << endl;
				o = path[o];
			
			cout << data[o] << " <- " << data[path[o]] << endl;
		
		else
		
			cout << "The road is death" << endl;
		
	


	

;

int main()

	Graph g;
	int s;
	cin >> s;
	g.dijkstra(s);
	int o;
	cin >> o;
	g.printPath(o);
	return 0;

测试如下: