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"]

];

我需要在单独的组中找到所有 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谷歌搜索了这个，这就是我如何了解到这被称为“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，但任何其他语言的示例都会非常有帮助。谢谢。

最佳答案

这可以使用广度优先搜索来解决。

这个想法是通过跳到相邻顶点来遍历源顶点的所有可达顶点。首先访问紧邻源顶点的顶点，然后是 2 跳远的顶点，依此类推。

由于使用的是边列表 图形表示，这里的代码效率不是很高。为了获得更好的性能，您可能需要使用邻接列表。

下面是一些可用的 JavaScript 代码。我使用 node.js 来运行这个:

// 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:我更新了我的答案以演示如何将边列表转换为邻接列表。该代码已注释，应该可以很好地说明这个概念。广度优先搜索功能被修改为使用邻接表，以及另一个轻微的修改(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);

关于javascript - 查找无向图的所有连通分量，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/21900713/