`Algorithm-Solution` `LeetCode` 6256. 将节点分成尽可能多的组

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Solution

Given a undirected unweighted graph $G$, if we divide it into $S1,S2,...$ groups, each group has a unique depth, that is, $Depth(Si)=i$, every point $a$ belongs to a unique group $S(a)$.
Then any edge between $a$ and $b$ satisfying $|Depth(S(a))-Depth(S(b))|=1$, or you can view groups as a layered-graph, where $S1$ at the top, all edges are connecting two adjacent groups.

Find the maximal number of group-count.

Firstly, because $G$ may contains multiple connected graph $Gi$, we can handle $Gi$ alone, then together them.
. For instance, if $G1$ has $3$ groups and $G2$ has $2$ groups, then $G$ has $5$ groups where the front $3$ groups denote $G1$, and so on.

Then, the question is, how many groups can a $Gi$ divides into at maximum (or it can’t be grouped (you can use Bipartite-Check, because if we let $L=S1,S3,S5,...$ and $R=S2,S4,...$, it must be a Bipartite. In fact, this case is not necessary handled by this way, it can be embedded into the below approach))
If $Gi$ can not be grouped, $G$ is also could not be grouped (unsolvable)
$Gi$ may be grouped in different ways (i.e., with different count of groups)
. One prominent property for $Gi$ is, if $Gi$ can be grouped, $S1$ (the first/top group) must can contains just one point. (if $S1=(a,...)$, $...$ can be putted into $S3$)
So, we choose a point $s$ as the start-point (the only point is $S1$), perform BFS_1 from $s$ with $depth[s]=1$, then if $depth[nex]\ne -1$, it must satisfies $|depth[nex]-depth[cur]|=1$ (otherwise, it is unsolvable), the deepest depth is the group-count. (e.g., 1-2-3, BFS(2) will get $2$, BFS(1/3) will get $3$, so $max(2,3,3)$ is the optimal scheme for $Gi$.)
Formally, let $fa$ be the Disjoint-Set root for $Gi$, each $BFS(i)$ with the deepest depth $d$ is used to update $Md[fa]=max(self,d)$ ($Md[fa]$ means the optimal group-division for $Gi$ which is represented by $fa$) (once $BFS(i)$ is failed, then $G$ is unsolvable)

Code

class Solution {
public:
    int D[ 502];
    int Md[ 502];
    int magnificentSets(int N, vector<vector<int>>& A) {
        Graph_UnWeighted g( N, A.size() * 2);
        g.Initialize( N);
        DisjointSet ss( N);
        ss.Initialize( N);
        for( auto & i : A){
            int a = i[0], b = i[1];
            -- a, -- b;
            g.Add_edge( a, b);
            g.Add_edge( b, a);
            ss.Merge( a, b);
        }
        memset( Md, 0, sizeof( Md));
        for( int i = 0; i < N; ++i){ 
            queue< int> que;
            que.push( i);
            memset( D, -1, sizeof( D));
            int aa = 1;
            D[ i] = 1;
            while( false == que.empty()){
                int cur = que.front();
                que.pop();
                for( int nex, e = g.Head[ cur]; ~e; e = g.Next[ e]){
                    nex = g.Vertex[ e];
                    if( D[ nex] == -1){
                        D[ nex] = D[ cur] + 1;
                        aa = max( aa, D[ nex]);
                        que.push( nex);
                    }
                    else{
                        if( ( D[ nex] != D[ cur] - 1) && ( D[ nex] != D[ cur] + 1)){
                            return -1;
                        }
                    }
                }
            }
            int fa = ss.Get_root( i);
            Md[ fa] = max( Md[ fa], aa);
        }
        int ans = 0;
        for( int i = 0; i < N; ++i){
            if( i == ss.Get_root( i)){
                ans += Md[ i];
            }
        }
        return ans;
    }
};
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