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不相交集数据结构

[英]disjoint set data structure

this is the code for findind disjoint sets 这是findind不交集的代码

    class disjoint_sets {
      struct disjoint_set {
      size_t parent;
      unsigned rank;
      disjoint_set(size_t i) : parent(i), rank(0) { }
       };
      std::vector<disjoint_set> forest;
       public:
       disjoint_sets(size_t n){
       forest.reserve(n);
        for (size_t i=0; i<n; i++)
         forest.push_back(disjoint_set(i));
        }
      size_t find(size_t i){
        if (forest[i].parent == i)
        return i;
         else {
        forest[i].parent = find(forest[i].parent);
        return forest[i].parent;
           }
          }
         void merge(size_t i, size_t j) {
          size_t root_i = find(i);
         size_t root_j = find(j);
          if (root_i != root_j) {
           if (forest[root_i].rank < forest[root_j].rank)
           forest[root_i].parent = root_j;
          else if (forest[root_i].rank > forest[root_j].rank)
          forest[root_j].parent = root_i;
           else {
            forest[root_i].parent = root_j;
             forest[root_j].rank += 1;
               }
            }
           }
            };

why are we incrementing rank if ranks are eqaul???nma beginner sorry and also what is the find step doing?? 如果等级为eqaul,为什么我们要增加等级?? nma初学者不好意思,查找步骤在做什么?

Because in this case - you add one tree is a "sub tree" of the other - which makes the original subtree increase its size. 因为在这种情况下-您添加的一棵树是另一棵树的“子树”-这会使原始子树增加其大小。

Have a look at the following example: 看下面的例子:

1           3
|           |
2           4

In the above, the "rank" of each tree is 2. 在上面,每棵树的“等级”是2。
Now, let's say 1 is going to be the new unified root, you will get the following tree: 现在,假设1将成为新的统一根,您将获得以下树:

    1
   / \
  /   \
 3     2
 |
 4

after the join the rank of "1" is 3, rank_old(1) + 1 - as expected. 加入后,“ 1”的等级为3, rank_old(1) + 1如预期。

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