深度优先搜索遍历连通图的递归算法

//深度优先搜索遍历连通图的递归算法
#include <iostream>
using namespace std;

#define MVNun 100
typedef char VerTexType;
typedef int ArcType;

typedef struct {
	VerTexType vexs[MVNun];
	ArcType arcs[MVNun][MVNun];
	int vexnum, arcnum;
}Graph;

bool visited[MVNun];
int FirstAdjVex(Graph G, int v);
int NextAdjVex(Graph G, int v, int w);

int LocateVex(Graph G, VerTexType v) {
	for (int i = 0;i < G.vexnum;++i) {
		if (G.vexs[i] == v)
			return i;
	}
	return -1;
}

void CreateUDN(Graph& G) {
	int i, j, k;
	cout << "请输入总顶点数，总边数 , 以空格隔开:";
	cin >> G.vexnum >> G.arcnum;
	cout << endl;
	cout << "输入点的名称，如 a：" << endl;

	for (i = 0;i < G.vexnum;++i) {
		cout << "请输入第" << (i + 1) << "请输入第";
		cin >> G.vexs[i];
	}
	cout << endl;

	for (i = 0;i < G.vexnum;++i)
		for (j = 0;j < G.vexnum;++j)
			G.arcs[i][j] = 0;
	cout << "输入边依附的顶点，如：a b" << endl;
	for (k = 0;k < G.arcnum;++k) {
		VerTexType v1, v2;
		cout << "请输入第" << (k + 1) << "条边依附的顶点:";
		cin >> v1 >> v2;
		i = LocateVex(G, v1);
		j = LocateVex(G, v2);
		G.arcs[j][i] = G.arcs[i][j] = 1;
	}
}

void DFS(Graph G, int v) {
	cout << G.vexs[v] << "    ";  visited[v] = true;
	int w;
	for (w = FirstAdjVex(G, v); w >= 0; w = NextAdjVex(G, v, w))
		if (!visited[w]) DFS(G, w);
}

int FirstAdjVex(Graph G, int v) {
	int i;
	for (i = 0;i < G.vexnum;++i) {
		if (G.arcs[v][i] == 1 && visited[i] == false)
			return i;
	}
	return -1;
}

int NextAdjVex(Graph G, int v, int w) {
	int i;
	for (i = w;i < G.vexnum;++i) {
		if (G.arcs[v][i] == 1 && visited[i] == false)
			return i;
	}
	return -1;
}

int main() {
	cout << "深度优先搜索遍历连通图的递归算法";
	Graph G;
	CreateUDN(G);
	cout << endl;
	cout << "无向连通图G创建完成！" << endl;

	cout << "请输入遍历连通图的起始点：";
	VerTexType c;
	cin >> c;

	int i;
	for (i = 0;i < G.vexnum;++i) {
		if (c == G.vexs[i])
			break;
	}
	cout << endl;
	while (i >= G.vexnum) {
		cout << "该点不存在，请重新输入！" << endl;
		cout << "请输入遍历连通图的起始点：";
		cin >> c;
		for (i = 0;i < G.vexnum;++i) {
			if (c == G.vexs[i]) {
				break;
			}
		}
	}
	cout << "深度优先搜索遍历连通图结果：" << endl;
	DFS(G, i);

	cout << endl;
	return 0;
}