有向图中

Question

我获得了一个定向图，并且在其中给出了两个节点，我需要找到可以从两个节点到达的最近的节点。唯一的问题是我能用两个dfs做到这一点，但我被告知要在O（logn）中做到这一点。附加限制是每个单元格可以具有最多一个传出边缘。输入以大小为N的数组给出，其中数组中的eachentry表示此节点所指向的节点的索引。所以这是我尝试的代码（不完全是dfs但仍然）：

int leastCommonDescendent(int nodes[], int N, int node1, int node2)
{
  int *visited = new int [N];
  int cnt1 = 0; //used for counting length of path from node1 to node2
  int cnt2 = 0; //used for counting length of path from node2 to node1
  int mark = node1; //storing as a marker needed later for detecting end of search

  if(node1 == node2) return node2;
  for(int i = 0; i < N; i++){
    visited[i] = 0;
  }

  while((nodes[node1] != node1) && (nodes[node1] != -1) && (visited[node1] == 0) && (node1 != node2)){
    visited[node1]++;
    node1 = nodes[node1];
    cnt1++;
  }

  visited[node1]++; //so that first node in cycle has count 2
                    //if find a node in 2nd iteration that has count 2
                    //such that when node1 == node2 it means we are in the same subgraph
                    //elsif node1 != node2 we are in different sub graphs

  while((nodes[node2] != node2) && (nodes[node2] != -1) && (visited[node2] != 2) && (node1 != node2)){
    visited[node2]++;
    node2 = nodes[node2];
    cnt2++;
  }
  //In below case the nodes are in different disjoint subgraphs
  //and both subgraphs have loops so node1 can never be equal to node2
  //cout << visited[node1] << visited[node2] << endl;
  if(node1 != node2) return -1;
    //In below case both nodes are in different disjoint subgraphs
    //but there is no loop in 1st one(containing node1)
    //and 2nd one has a loop
  if ((nodes[node1] == -1) && (visited[node2] == 1)) return -1;
    //In below case both nodes are in different disjoint subgraphs
    //but 1st one has a loop and second one doesn't
  if(nodes[node2] == -1) return -1;
    //In below case both nodes are in same subgraph so we
    //need to check the length of two alternate paths
  if(cnt1 > cnt2)
    return node2;
  else
    return mark;
}  

假设我有一个图形组件（基本上是一个子图），如下所示：

在这种情况下，如果我想从7＆amp;找到最近的节点9我得到答案9虽然它应该是8.虽然我明白这是因为我在两个循环中都有条件cell1！= cell2但是我要去整个循环以防我删除那些邀请更多时间。此外，我觉得这个解决方案混杂了多个if。我们能有更简单的解决方案吗？ （可能 O（logn））

此图也可以具有如上图所示的周期。因此，我猜不可能转换为树。

Answer 1

通过简单地反转图表的链接，可以很容易地将其缩小为（和来自）Lowest Common Ancestor in a tree（或准确地说是在森林中）。

一般情况下，可以在O(h)中完成，步进＆＃34; up＆＃34;树一步一步（在原始图中前进），并将找到的节点存储在一个集合中，直到设置的相交不为空。

如果允许预处理，可以在线性时间预处理它以获得更好的弹性。