递归 + 标记
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;
}
