CSES Solutions - Tree Matching

Last Updated : 03 May, 2024

You are given a tree consisting of n nodes. A matching is a set of edges where each node is an endpoint of at most one edge. What is the maximum number of edges in a matching ?

Examples:

Input : N = 10 , edges = {{5,8}, {4,6}, {9,1}, {10,4}, {1,3}, {2,3}, {7,9}, {6,2}, {5,10}}
Output : 5
Explanation : One possible matching is {5-8 , 10-4, 1-3, 7-9 , 6-2}

Input : N = 10 , edges = {{6,3}, {2,8}, {1,8}, {4,10}, {7,3}, {8,7}, {1,5}, {5,9}, {10,9}}
Output : 4
Explanation : one possible matching is {6-3, 2-8, 4-10, 1-5}

Approach:

We can solve this problem using Dynamic Programming (DP) with a 2D dp array to track maximum "pairs" achievable in subtrees. There are two conditions:

Exclude current node: dp[0][u] stores the maximum pairs possible without including node u itself.
Include current node: dp[1][u] stores the maximum pairs possible by including u in a matching with a child.
A Depth-First Search (DFS) calculates optimal values for dp at each node based on its children. Finally, the maximum number of pairs in a perfect matching is the maximum between dp[0][0] (excluding root) and dp[1][0] (including root).

Step by Step Algorithm:

Initialization: For each node u, initialize dp[0][u] to 0 and dp[1][u] to negative infinity.
DFS (Depth-First Search): Perform a DFS traversal on the tree.
For each child v of node u:
- Recursively call DFS on child node v.
- Update dp[0][u]: Add the maximum of dp[1][v] (pairs including v) and dp[0][v] (pairs excluding v) to dp[0][u]. This represents the maximum pairs possible in the subtree rooted at u without including u.
- Update dp[1][u]: Calculate the difference between dp[0][v] and dp[1][v]. This represents the maximum number of pairs gained by including u in the matching with child v compared to the case where v is not included. Choose the maximum between the current dp[1][u] and this difference. This ensures we select the scenario that leads to the most pairs when including u.

Final Update: After processing all children, update dp[1][u] by adding dp[0][u]. This incorporates the pairs possible without including u to the total pairs when u is included.
Output: After the DFS traversal, the maximum number of pairs is the maximum value between dp[0][0] (excluding root) and dp[1][0] (including root).

Below is the implementation of the above algorithm:

C++

#include <algorithm>
#include <iostream>
#include <vector>

using namespace std;
#define ll long long
// Represents infinity
const ll INF = 10000000000000;
// Maximum number of nodes
const ll MX = 2e5 + 5;

// Adjacency list
vector<ll> Adj[MX];
// Dynamic programming array
ll dp[2][MX];

// Depth-first search to compute the maximum number of pairs
// u: current node
// p: parent node
void DFS(ll u, ll p)
{
    // Initialize dp values
    // Maximum number of pairs without including the current
    // node
    dp[0][u] = 0;
    // Maximum number of pairs including the current node
    dp[1][u] = -INF;

    // Traverse through each adjacent node
    for (ll v : Adj[u]) {
        if (v == p)
            // Skip if it's the parent node
            continue;

        // Recursively call DFS for child node
        DFS(v, u);

        // Update dp values
        dp[0][u] += max(
            dp[1][v],
            // Add maximum pairs without including the child
            dp[0][v]);
        ll opt = min(dp[0][v] - dp[1][v],
                     // Difference between including and
                     // excluding the child
                     (long long)0);
        // Update maximum pairs including the current node
        dp[1][u] = max(dp[1][u], opt);
    }

    // Update dp[1] value
    // Add pairs without the current node
    dp[1][u] += dp[0][u];
    // Add the current node to the matching
    dp[1][u]++;
}

int main()
{
    // Number of nodes
    ll n = 10;

    // Edges of the tree
    ll edges[][2] = { { 5, 8 },  { 4, 6 }, { 9, 1 },
                      { 10, 4 }, { 1, 3 }, { 2, 3 },
                      { 7, 9 },  { 6, 2 }, { 5, 10 } };

    // Build adjacency list
    for (ll i = 0; i < n - 1; ++i) {
        // Convert to 0-based indexing
        ll a = edges[i][0] - 1;
        // Convert to 0-based indexing
        ll b = edges[i][1] - 1;
        Adj[a].push_back(b);
        // Undirected tree, add edge in both directions
        Adj[b].push_back(a);
    }

    // Start DFS from root node (0)
    DFS(0, -1);

    // Output the maximum number of pairs
    ll ans = max(dp[0][0], dp[1][0]);
    cout << "Maximum number of edges in a matching: " << ans
         << endl;

    return 0;
}

Java

import java.util.ArrayList;

public class MaximumMatchingPairs {
    // Represents infinity
    static final long INF = 10000000000000L;
    // Maximum number of nodes
    static final int MX = 200005;

    // Adjacency list
    static ArrayList<Integer>[] Adj = new ArrayList[MX];
    // Dynamic programming array
    static long[][] dp = new long[2][MX];

    // Depth-first search to compute the maximum number of
    // pairs u: current node p: parent node
    static void DFS(int u, int p)
    {
        // Initialize dp values
        // Maximum number of pairs without including the
        // current node
        dp[0][u] = 0;
        // Maximum number of pairs including the current
        // node
        dp[1][u] = -INF;

        // Traverse through each adjacent node
        for (int v : Adj[u]) {
            if (v == p)
                // Skip if it's the parent node
                continue;

            // Recursively call DFS for child node
            DFS(v, u);

            // Update dp values
            dp[0][u] += Math.max(dp[1][v], dp[0][v]);
            long opt = Math.min(dp[0][v] - dp[1][v], 0);
            // Update maximum pairs including the current
            // node
            dp[1][u] = Math.max(dp[1][u], opt);
        }

        // Update dp[1] value
        // Add pairs without the current node
        dp[1][u] += dp[0][u];
        // Add the current node to the matching
        dp[1][u]++;
    }

    public static void main(String[] args)
    {
        // Number of nodes
        int n = 10;
        // Edges of the tree
        int[][] edges = { { 5, 8 },  { 4, 6 }, { 9, 1 },
                          { 10, 4 }, { 1, 3 }, { 2, 3 },
                          { 7, 9 },  { 6, 2 }, { 5, 10 } };

        for (int i = 0; i < MX; i++) {
            Adj[i] = new ArrayList<>();
        }

        // Build adjacency list
        for (int i = 0; i < n - 1; ++i) {
            int a = edges[i][0] - 1;
            int b = edges[i][1] - 1;
            Adj[a].add(b);
            Adj[b].add(a);
        }

        // Start DFS from root node (0)
        DFS(0, -1);

        // Output the maximum number of pairs
        long ans = Math.max(dp[0][0], dp[1][0]);
        System.out.println(
            "Maximum number of edges in a matching: "
            + ans);
    }
}

// This code is contributed by shivamgupta310570

Python

# Represents infinity
INF = 10000000000000
# Maximum number of nodes
MX = 200005

# Adjacency list
Adj = [[] for _ in range(MX)]
# Dynamic programming array
dp = [[0] * MX for _ in range(2)]

# Depth-first search to compute the maximum number of pairs
# u: current node
# p: parent node


def DFS(u, p):
    # Initialize dp values
    # Maximum number of pairs without including the current node
    dp[0][u] = 0
    # Maximum number of pairs including the current node
    dp[1][u] = -INF

    # Traverse through each adjacent node
    for v in Adj[u]:
        if v == p:
            # Skip if it's the parent node
            continue

        # Recursively call DFS for child node
        DFS(v, u)

        # Update dp values
        dp[0][u] += max(dp[1][v], dp[0][v])
        opt = min(dp[0][v] - dp[1][v], 0)
        # Update maximum pairs including the current node
        dp[1][u] = max(dp[1][u], opt)

    # Update dp[1] value
    # Add pairs without the current node
    dp[1][u] += dp[0][u]
    # Add the current node to the matching
    dp[1][u] += 1


# Number of nodes
n = 10

# Edges of the tree
edges = [[5, 8], [4, 6], [9, 1], [10, 4], [
    1, 3], [2, 3], [7, 9], [6, 2], [5, 10]]

# Build adjacency list
for edge in edges:
    # Convert to 0-based indexing
    a = edge[0] - 1
    # Convert to 0-based indexing
    b = edge[1] - 1
    Adj[a].append(b)
    # Undirected tree, add edge in both directions
    Adj[b].append(a)

# Start DFS from root node (0)
DFS(0, -1)

# Output the maximum number of pairs
ans = max(dp[0][0], dp[1][0])
print("Maximum number of edges in a matching:", ans)

JavaScript

// Represents infinity
const INF = 10000000000000;
// Maximum number of nodes
const MX = 2e5 + 5;

// Adjacency list
let Adj = new Array(MX).fill().map(() => []);
// Dynamic programming array
let dp = new Array(2).fill().map(() => new Array(MX).fill(0));

// Depth-first search to compute the maximum number of pairs
// u: current node
// p: parent node
function DFS(u, p) {
    // Initialize dp values
    // Maximum number of pairs without including the current node
    dp[0][u] = 0;
    // Maximum number of pairs including the current node
    dp[1][u] = -INF;

    // Traverse through each adjacent node
    for (let v of Adj[u]) {
        if (v === p)
            // Skip if it's the parent node
            continue;

        // Recursively call DFS for child node
        DFS(v, u);

        // Update dp values
        dp[0][u] += Math.max(
            dp[1][v],
            // Add maximum pairs without including the child
            dp[0][v]);
        let opt = Math.min(dp[0][v] - dp[1][v], 0);
        // Update maximum pairs including the current node
        dp[1][u] = Math.max(dp[1][u], opt);
    }

    // Update dp[1] value
    // Add pairs without the current node
    dp[1][u] += dp[0][u];
    // Add the current node to the matching
    dp[1][u]++;
}

// Number of nodes
let n = 10;

// Edges of the tree
let edges = [ [ 5, 8 ], [ 4, 6 ], [ 9, 1 ],
              [ 10, 4 ], [ 1, 3 ], [ 2, 3 ],
              [ 7, 9 ], [ 6, 2 ], [ 5, 10 ] ];

// Build adjacency list
for (let i = 0; i < n - 1; ++i) {
    // Convert to 0-based indexing
    let a = edges[i][0] - 1;
    // Convert to 0-based indexing
    let b = edges[i][1] - 1;
    Adj[a].push(b);
    // Undirected tree, add edge in both directions
    Adj[b].push(a);
}

// Start DFS from root node (0)
DFS(0, -1);

// Output the maximum number of pairs
let ans = Math.max(dp[0][0], dp[1][0]);
console.log("Maximum number of edges in a matching:", ans);

Output

Maximum number of edges in a matching: 5

Time Complexity : O(V)
Space Complexity : O(V)

CSES Solutions - Tree Matching

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CSES Solutions - Tree Matching

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