查找无向图的所有连通分量

Question

我有一个对象列表(无向边),如下所示:

pairs = [

 pair:["a2", "a5"],
 pair:["a3", "a6"],
 pair:["a4", "a5"],
 pair:["a7", "a9"]

];

我需要在不同的组中找到所有组件(连接的节点).所以从给定的对中我需要得到:

groups = [
  group1: ["a2", "a5", "a4"],
  group2: ["a3", "a6"],
  group3: ["a7", "a9"]
];

我实际上在这里阅读了一些答案并用Google搜索,这就是我学习它的方法,称为"在图中查找连接的组件",但是找不到任何示例代码.我在Node.js上使用JavaScript,但任何其他语言的样本都会非常有用.谢谢.

Answer 1

这可以使用广度优先搜索来解决.

我们的想法是通过跳转到相邻的顶点来遍历源顶点的所有可到达顶点.首先访问源顶点旁边的顶点,然后是2跳的顶点,等等.

由于使用了图表表示,这里的代码效率不高,这是一个边缘列表.要获得更好的性能,您可能需要使用邻接列表.

这是JavaScript中的一些工作代码.我曾经node.js这样做:

// Breadth First Search function
// v is the source vertex
// all_pairs is the input array, which contains length 2 arrays
// visited is a dictionary for keeping track of whether a node is visited
var bfs = function(v, all_pairs, visited) {
  var q = [];
  var current_group = [];
  var i, nextVertex, pair;
  var length_all_pairs = all_pairs.length;
  q.push(v);
  while (q.length > 0) {
    v = q.shift();
    if (!visited[v]) {
      visited[v] = true;
      current_group.push(v);
      // go through the input array to find vertices that are
      // directly adjacent to the current vertex, and put them
      // onto the queue
      for (i = 0; i < length_all_pairs; i += 1) {
        pair = all_pairs[i];
        if (pair[0] === v && !visited[pair[1]]) {
          q.push(pair[1]);
        } else if (pair[1] === v && !visited[pair[0]]) {
          q.push(pair[0]);
        }
      }
    }
  }
  // return everything in the current "group"
  return current_group;
};

var pairs = [
  ["a2", "a5"],
  ["a3", "a6"],
  ["a4", "a5"],
  ["a7", "a9"]
];

var groups = [];
var i, k, length, u, v, src, current_pair;
var visited = {};

// main loop - find any unvisited vertex from the input array and
// treat it as the source, then perform a breadth first search from
// it. All vertices visited from this search belong to the same group
for (i = 0, length = pairs.length; i < length; i += 1) {
  current_pair = pairs[i];
  u = current_pair[0];
  v = current_pair[1];
  src = null;
  if (!visited[u]) {
    src = u;
  } else if (!visited[v]) {
    src = v;
  }
  if (src) {
    // there is an unvisited vertex in this pair.
    // perform a breadth first search, and push the resulting
    // group onto the list of all groups
    groups.push(bfs(src, pairs, visited));
  }
}

// show groups
console.log(groups);

更新:我已更新我的答案,演示如何将边缘列表转换为邻接列表.代码被评论,应该很好地说明这个概念.修改宽度优先搜索功能以使用邻接列表,以及另一个略微修改(关于将顶点标记为已访问).

// Converts an edgelist to an adjacency list representation
// In this program, we use a dictionary as an adjacency list,
// where each key is a vertex, and each value is a list of all
// vertices adjacent to that vertex
var convert_edgelist_to_adjlist = function(edgelist) {
  var adjlist = {};
  var i, len, pair, u, v;
  for (i = 0, len = edgelist.length; i < len; i += 1) {
    pair = edgelist[i];
    u = pair[0];
    v = pair[1];
    if (adjlist[u]) {
      // append vertex v to edgelist of vertex u
      adjlist[u].push(v);
    } else {
      // vertex u is not in adjlist, create new adjacency list for it
      adjlist[u] = [v];
    }
    if (adjlist[v]) {
      adjlist[v].push(u);
    } else {
      adjlist[v] = [u];
    }
  }
  return adjlist;
};

// Breadth First Search using adjacency list
var bfs = function(v, adjlist, visited) {
  var q = [];
  var current_group = [];
  var i, len, adjV, nextVertex;
  q.push(v);
  visited[v] = true;
  while (q.length > 0) {
    v = q.shift();
    current_group.push(v);
    // Go through adjacency list of vertex v, and push any unvisited
    // vertex onto the queue.
    // This is more efficient than our earlier approach of going
    // through an edge list.
    adjV = adjlist[v];
    for (i = 0, len = adjV.length; i < len; i += 1) {
      nextVertex = adjV[i];
      if (!visited[nextVertex]) {
        q.push(nextVertex);
        visited[nextVertex] = true;
      }
    }
  }
  return current_group;
};

var pairs = [
  ["a2", "a5"],
  ["a3", "a6"],
  ["a4", "a5"],
  ["a7", "a9"]
];

var groups = [];
var visited = {};
var v;

// this should look like:
// {
//   "a2": ["a5"],
//   "a3": ["a6"],
//   "a4": ["a5"],
//   "a5": ["a2", "a4"],
//   "a6": ["a3"],
//   "a7": ["a9"],
//   "a9": ["a7"]
// }
var adjlist = convert_edgelist_to_adjlist(pairs);

for (v in adjlist) {
  if (adjlist.hasOwnProperty(v) && !visited[v]) {
    groups.push(bfs(v, adjlist, visited));
  }
}

console.log(groups);

Answer 2

使用Disjoint Set Forest算法最好地解决您要解决的任务.

基本上它意味着以最佳方式准确解决您描述的问题.

• 为每个顶点分配它所属的顶点集.我们将其表示为root(v).在整个算法中,每个这样的集合将被视为种植树.因此,我们将与root(v)树的根相关联.所以我们假设root(v)返回一个顶点索引.
• 如果在整个算法中保留一个辅助数组 - 其种植树中每个顶点的父顶点,则算法将变得最简单.如果不存在这样的顶点,我们将存储-1而不是表示.因此,您parent[v] = -1为每个算法启动算法,v因为所有顶点最初都在自己的树中,而您没有父项
• 我们还需要一个额外的数组来存储一些被称为顶点等级的数组.它基本上等于以顶点为根的子树的深度.你不必太担心它.该数组用于性能优化.让我们说出来rank.我们将所有等级初始化为1
• 您开始逐个处理边缘,每个边缘可能会触发树的合并.这是您可以优化的有趣部分.

这是一个伪代码:

parent[v] = -1 for v in Vertices
rank[v] = 1 for v in Vertices
root (v):
   processed = []
   while parent[v] != -1
      processed << v
      v = parent[v]
   for vertex : processed
      parent = v // optimisation: here we move the assoc. trees to be directly connected the root
   return v

join (v1, v2):
  if rank[v1] < rank[v2]:
     parent[v1] = v2
  if rank[v1] > rank[v2]:
     parent[v2] = v1
  parent[v2] = v1
  rank[v1]++

merge_trees (v1, v2)
  root1 = root(v1)
  root2 = root(v2)
  if root1 == root2:
    // already in same tree nothing else to be done here
    return true
  else
    // join trees
    join (v1, v2)
    return false

main:
   numberTrees = size(Vertives)
   for edge: edges
      if merge_trees(edge.begin, edge.end):
         numberTrees--
   print numberTrees // this is the number you are interested in.

注意如果你没有太多的表现,你可以省略等级的事情.没有它,你的算法可能会运行得更慢,但也许你更容易理解和维护.在这种情况下,您可以join在任何方向上进行顶点.我应警告您,然后会存在会触发您的算法运行速度较慢的场景.