DFS(邻接矩阵表示)

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
using namespace std;

//#define PATH
#define VERTEX_NUM 8

#ifdef PATH
int count = 1;
#endif
bool visited[VERTEX_NUM + 1];        // 访问标志数组(备忘表)

int FirstAdjVex(bool G[VERTEX_NUM + 1][VERTEX_NUM + 1], int v)
{
    for (int j = 1; j <= VERTEX_NUM; j++) {
        if (G[v][j] == 1)
            return j;
    }
    cout << "该点是孤立点" << endl;    // 连通图则不会到这一步
    return -1;
}

int NextAdjVex(bool G[VERTEX_NUM + 1][VERTEX_NUM + 1], int v, int w)
{
    for (int j = w + 1; j <= VERTEX_NUM; j++) {
        if (G[v][j] == 1)
            return j;
    }
    return -1;
}


void DFS(bool G[VERTEX_NUM + 1][VERTEX_NUM + 1], int v)
{
    // 从第v个顶点出发递归地深度优先遍历图G。
    int w;
    visited[v] = true;
    cout << v << endl;
    for (w = FirstAdjVex(G, v);  w != -1;  w = NextAdjVex(G, v, w)) {
#ifdef PATH
        cout << "------------------" << count++ << "----" << v << "->" << w << endl;    // 1
#endif
        if (!visited[w]) {        // 对v的尚未访问的邻接顶点w递归调用DFS
            DFS(G, w);
#ifdef PATH
            cout << v << endl;    // 有待研究?                                            // 2
#endif
        }
    }
#ifdef PATH
    cout << "回溯路径" << v << "->";                                                    // 3
#endif
}

void DFSTraverse(bool G[VERTEX_NUM + 1][VERTEX_NUM + 1])
{
    // 对图G作深度优先遍历。
    int v;
    for (v = 1; v <= VERTEX_NUM; ++v)
        visited[v] = false;                    // 访问标志数组初始化
    for (v = 1; v <= VERTEX_NUM; ++v) {        // 若是连通图,则v=1即全遍历到
        if (!visited[v])
            DFS(G, v);                        // 对尚未访问的顶点调用DFS
    }
}

int main(int argc, char **argv)
{
    freopen("cin.txt", "r", stdin);
    bool G[VERTEX_NUM + 1][VERTEX_NUM + 1] = {0};
    int a;
    int b;
    while (cin >> a >> b, !(a == 0 && b == 0)) {    //以0 0作为输入结束
        G[a][b] = 1;
        G[b][a] = 1;
    }
    for (int i = 1; i <= VERTEX_NUM; i++) {            // 输出邻接矩阵
        for (int j = 1; j <= VERTEX_NUM; j++) {
            cout << G[i][j] << ' ';
        }
        cout << endl;
    }

    DFSTraverse(G);
    cout << endl;

    return 0;
}

/* cin.txt:
1 2
2 4
2 5
8 4
8 5
1 3
3 6
3 7
6 7
0 0
*/

 

原文地址:https://www.cnblogs.com/jjtx/p/2550042.html