从JUNG图表创建邻接矩阵

Question

Graph < Integer, Integer> g = new SparseMultigraph<Integer, Integer>();
    g.addVertex(1);g.addVertex(2);g.addVertex(3);
    g.addEdge(0,1,2 ,EdgeType.DIRECTED);g.addEdge(1,2,3 ,EdgeType.DIRECTED);g.addEdge(2,3,1 ,EdgeType.DIRECTED);g.addEdge(3,1,3 ,EdgeType.DIRECTED);

如何转换成考虑到，这是一个有向图的邻接矩阵这个曲线图。

Answer 1

在此信息可以找到一个邻接矩阵：

宽度与深度优先搜索 - 第3部分

如何实现它？

// Adjacency matrix
int map[21][21] = {

/*   A B C D E F G H I L M N O P R S T U V Z */
  {0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9,0},
  {1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,1}, // Arad
  {2,0,0,0,0,0,1,1,0,0,0,0,0,0,1,0,0,0,1,0,0}, // Bucharest
  {3,0,0,0,1,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0}, // Craiova
  {4,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0}, // Dobreta
  {5,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0}, // Eforie
  {6,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0}, // Fagaras
  {7,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, // Girgiu
  {8,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0}, // Hirsova
  {9,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0}, // Iasi
  {0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0}, // Lugoj
  {1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0}, // Mehadia
  {2,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0}, // Neamt
  {3,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1}, // Oradea
  {4,0,1,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0}, // Pitesti
  {5,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0}, // Rimnicu Vilcea
  {6,1,0,0,0,0,1,0,0,0,0,0,0,1,0,1,0,0,0,0,0}, // Sibiu
  {7,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0}, // Timisoara
  {8,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0}, // Urziceni
  {9,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0}, // Vaslui
  {0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0}  // Zerind
};

注意，第一注释行表示各城市的名称的首字母。与邻接矩阵进行映射是指这些信件，使其更容易理解。例如，获取引用阿拉德邻接矩阵的第一个条目：我们有阿拉德有导致我们锡比乌，蒂米什瓦拉和Zerind路径，因此，我们把1的值上，代表这些城市的列，在这种情况下， ，字母S，T和Z的下方列那怎么映射完成。我们把国家的0值的其他列，没有路径导致我们那些城市。

鉴于你图，迭代其边缘并创建邻接矩阵。