kruskal.cpp

用Kruskal算法实现若干个城市之间的最短路径.最大城市数目为7个.

// Kruskal.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include "adjMList.h"
#include "iostream.h"
#include<iomanip.h>
#include<stdlib.h>


int main(int argc, char* argv[])
{
	int V,E;         //定义图的点数及边数等
	RCW rcw[21]={{0,1,50},{0,2,60},{1,2,39},{0,3,22},{1,3,65},{2,3,52},
				{0,4,44},{1,4,40},{2,4,37},{3,4,51},{0,5,55},{1,5,80},
				{2,5,59},{3,5,30},{4,5,70},{0,6,66},{1,6,62},{2,6,45},
				{3,6,42},{4,6,49},{5,6,72}};
	cout<<"警告!! 输入城镇数不得超过7个"<<endl;
	cout<<"请输入城镇数：V=";cin>>V;
	E=(V*(V-1))/2;
	cout<<"各城镇间共有"<<E<<"条可达路线"<<endl;
	static edge AE[21];        //定义边集数组
	adjMList B(AE,V,E);
	B.CreateMatrix(V,E,rcw);         //由图的邻接矩阵生成图的边集数组
	B.ChangeEdgeSet(AE,V,E);
	cout<<"城镇连通图的邻接矩阵生成的边集数组为:"<<endl;
	B.OutputEdgeSet(AE,E);
	cout<<"各城镇间的最短连接路径为:"<<endl;
	B.Kruskal(AE,V,E);
	return 0;
}