图的遍历——DFS(邻接矩阵)

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一个连通图只要DFS一次,即可打印所有的点。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <malloc.h>

using namespace std;

const int VERTEX_NUM = 20;
const int INFINITY = 0x7fffffff; 		// 最大int型数,表示权的无限值 

bool vis[VERTEX_NUM];

class Graph {
public:
	int vexNum;
	int edgeNum;
	int vex[VERTEX_NUM];
	int arc[VERTEX_NUM][VERTEX_NUM]; 
};

void createGraph(Graph &G)
{
	cout << "please input vexNum and edgeNum: ";
	cin >> G.vexNum >> G.edgeNum;
	for (int i = 0; i != G.vexNum; ++i) {
		cout << "please input no" << i+1 << " vertex: ";
		cin >>  G.vex[i];
	}
	for (int i = 0; i != G.vexNum; ++i) {
		for (int j = 0; j != G.vexNum; ++j) {
			G.arc[i][j] = INFINITY;
		}
	}
	for (int k = 0; k != G.edgeNum; ++k) {
		cout << "please input the vertex of edge(vi, vj) and weight: ";
		int i, j, w;
		cin >> i >> j >> w;
		G.arc[i][j] = w;
		G.arc[j][i] = G.arc[i][j];			// 无向图 
	}
}

void DFS(const Graph &G, int k)
{
	vis[k] = true;
	cout << G.vex[k] << " ";			// 打印图的结点 
	for (int i = 0; i != G.vexNum; ++i) {
		if (G.arc[k][i] != INFINITY && !vis[i]) 
			DFS(G, i);
	}
}

void DFSTraverse(const Graph &G)
{
	memset(vis, false, VERTEX_NUM);
	// 连通图一次即遍历完成 
	for (int i = 0; i != G.vexNum; ++i) {
		if (!vis[i])
			DFS(G, i);
	}
} 

int main()
{
	Graph G;
	createGraph(G);
	DFSTraverse(G);
	return 0;
} 

  

posted @ 2018-03-25 19:50  GGBeng  阅读(555)  评论(0编辑  收藏  举报