PROBLEM LINK  Author  Sundar Annamalai DIFFICULTY – Medium PREREQUISITES – PROBLEM IN SHORT  Given a tree of size $N$ rooted at 1 and $M$ lists of integers. $A_i$ denotes $i_{th}$ list. Consider a bipartite graph $G$, having $M$ nodes in left bipartition(denoted by 'L') and N nodes in right one(denoted by 'R'). There is an edge from $L_i$ to all $R_{descendants(j)}$ if $A_i$ contains $j$. Find the maximum matching in the graph. QUICK EXPLANATION – We will construct a flow network. We will add an edge from source to each $L_i$ (left bipartition, contains $M$ nodes), each of capacity one. We will add an edge from $L_i$ to $R_j$, each of capacity one, if $A_i$ contains $R_j$ as its element. From each $R_j$, we will add an edge to $R_{child(j)}$ with the capacity of AT LEAST size of subtree of $child(j)$. We add an edge from each $R_j$ to sink of capacity one. Maxflow of this network is our required answer. EXPLANATION – I will use $‘L’$ for denoting left bipartition, containing $M$ nodes and $‘R’$ for denoting right partition, containing $N$ nodes. Let us start with the most naive way of solving this problem which is, to actually construct the graph $G$, and find the size of maximum matching in that graph using Maxflow algorithm. This reduces to a classic problem and can be done easily. The Bipartite graph $G$ can be constructed by adding an edge from $L_i$ to all $R_{descendants(j)}$ (including $R_j$) of capacity one, where $A_i$ contains $j$. TIME COMPLEXITY – $O(M * N)$ – For adding edges Since our Maxflow algorithm can’t run in time limit, we need to reduce the number of edges in our graph. KEY OBSERVATIONS – $1.$ If there is an edge from $L_i$ to some $R_j$, we will also have an edge from $L_i$ to $R_k$, where $k$ is any descendant of $j$. A more explained example  INTENDED SOLUTION – With the help of above observations, we can say that instead of adding edges from each $L_i$ to all $R_{descendants(j)}$, where $A_i$ contains $j$, we will now add edge from $L_i$ to only elements of $A_i$ and from each $R_j$ to all $R_k$, where $k$ is a descendant of $j$. All edges are of capacity one. This doesn’t help the time complexity, but the problem can be reduced from this stage. Now, we again apply second observation to say that instead of direct flow from each $R_j$ to all $R_k$, where $k$ is a descendant of $j$, we make the path of flow from $j$ to $k$, go via all the $R_p$, where $p$ is a node which lies between the path from $j$ to $k$ in the tree. $p$ can be defined as child of $j$ and ancestor of $k$ in the tree. When we do this, we will have sizeofsubtree($p$) number of edges from $R_{parent(p)}$ to $R_p$. This is equivalent to a single edge from $R_{parent(p)}$ to $R_p$ of capacity of size of subtree of $p$. Replacing $p$ by $j$, all we need to do is only add edges from each $R_{parent(j)}$ to $R_j$ with the capacity of size of subtree of $j$ and run the maxflow algorithm. Another way to look upon this is to use observation 1 and say that if there is an edge from $R_j$ to some $R_k$, $R_j$ will also have an edge to all $R_{descendants(k)}$, so we can say it is enough to have an edge from all $R_{parent(c)}$ to $R_c$ with the capacity of at least size of subtree of $c$ for some c to keep propagating the flow towards all the descendants. I used the word ‘at least’ because we are already keeping the capacity of the edge such that at least one unit is reaching every descendant, and because every $R_j$ can send only one unit to the sink, it won’t matter if we keep the capacity to be size of subtree or one trillion. So, you don't have to actually find size of subtree of every node but instead, keep the capacities of the edges greater than size of tree$(N)$. PROOF OF CORRECTNESS  We can say that given a matching in our original bipartite graph $G$, you can get a flow of at least that much in this network. That's trivial, hence $matching <= flow$. And given a flow in our network, we can get a matching related to it in bipartite graph $G$. In our network, a flow path can be described as  $source, L_i, R_j, R_{child(j)}, ... , R_{K^{th} child(j)}, sink$. This path of flow will relate to the edge from $L_i$ to $R_{K^{th} child(j)}$ in bipartite graph $G$. Edges from source to $L_i$ and from $R_{K^{th} child(j)}$ to sink are used and thus, neither of both will be used again and each flow path corresponds to a unique edge between a unique $L_i$ and a unique $R_j$ in a matching. Thus, $flow <= matching$ $\implies$ $flow = matching$. Maxflow of this network will give maximum number of unique pairs and hence maximum matching in the bipartite graph $G$. $K$ can be zero also, in that case, it refers to itself. Each node is $0^{th}$ child of itself. TIME COMPLEXITY – Number of edges  $$=\hspace{1mm} \sum_{i=1}^{m}A_i\hspace{1mm}+\hspace{1mm}2*N\hspace{1mm}+\hspace{1mm}M$$ $$= \hspace{1mm}10000\hspace{1mm} +\hspace{1mm} 10000\hspace{1mm} +\hspace{1mm} 5000\hspace{1mm} =\hspace{1mm} 25000.$$ Maxflow can be at most $N = 5000$ Time complexity of Maxflow algorithm $= O(Edges * Maxflow) = O(1.25 * 10^8)$ per query. 4 queries should work in given time limit of 3.5 seconds. SECOND APPROACH  In this solution, instead of adding edge from $R_{parent(j)}$ to $R_j$ with the capacity of at least $subtree(j)$, we will add edge from $R_j$ to $R_{parent(j)}$ with the capacity of 'exactly' size of $subtree(j)$ and add only one edge from $R_1$ to the sink of capacity N. No need to add edges from other $R_j$ to sink. This works because, we just changed and redirected the flow from every $R_j$ to go via $R_1$ and added a single edge of capacity $N$ to the sink to account for the redirection of flow. This is again, through observation 2. This time, I used the word 'exactly' because, the edge from $R_1$ to sink is no longer of capacity one, and having more capacity can cause more flow and distort the answer. TIME COMPLEXITY  Second approach has a slightly higher time complexity by a constant factor than first one because the flow to sink can happen only from one node. Also, the edges have a limited capacity. All this increases the length of augmenting path and hence the time of DFS part in maxflow algorithm. No major change as such. IMPLEMENTATION  Editorialist's code for first approach  here. NOTE  This link has an excellent figure with a short explanation of the first approach for better understanding  here.
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asked 27 Jul, 00:45
