Overview

This problem asks us to find the shortest tour that traverses every edge at least once and returns to the starting vertex (the Chinese Postman Problem). We solve it by exploiting the constraint that there are at most 16 vertices of odd degree, using bitmask DP to compute a minimum weight perfect matching.

Analysis

Relationship with Eulerian Circuits

A tour that traverses every edge exactly once and returns to the starting vertex is called an Eulerian circuit. The condition for an Eulerian circuit to exist is “the graph is connected and every vertex has even degree.”

If the graph satisfies this condition, the answer is simply the sum of all edge weights \(\sum W_i\). There is no need to traverse any extra edges.

When Odd-Degree Vertices Exist

If there are vertices with odd degree, an Eulerian circuit does not exist. In this case, we need to traverse some edges two or more times (paying additional cost) to make all vertex degrees even, thereby enabling an Eulerian circuit.

A fundamental property of graph theory is that the number of odd-degree vertices is always even. By pairing up the odd-degree vertices and additionally traversing the edges along the shortest path for each pair, we can flip the parity of the degrees at both endpoints of that pair.

Minimizing the Cost

We need to find a perfect matching (partitioning all odd-degree vertices into pairs) of the odd-degree vertices such that the total shortest distance between each pair is minimized. This gives the minimum additional traversal cost.

Why Bitmask DP Works

Since the number of odd-degree vertices is at most 16, we can efficiently compute the minimum weight perfect matching using bitmask DP with \(2^{16} = 65536\) states. Naively enumerating all pairings would yield \((15)!! = 2027025\) combinations, but bitmask DP eliminates redundant computations.

Algorithm

1. Read the graph, and compute the total sum of all edge weights \(S = \sum W_i\) and the degree of each vertex.

2. Enumerate the odd-degree vertices. If there are 0 such vertices, an Eulerian circuit exists, so output \(S\) and terminate.

3. Compute shortest distances between odd-degree vertices. Run Dijkstra’s algorithm from each odd-degree vertex as the source, and record the shortest distances to all other odd-degree vertices (at most 16 runs of Dijkstra).

4. Find the minimum weight perfect matching using bitmask DP.

   • \(dp[\text{mask}]\) = the minimum cost of pairing up all vertices represented by \(\text{mask}\)
   • Starting from \(dp[0] = 0\), for each state, select the smallest-numbered unmatched vertex and try pairing it with each remaining unmatched vertex.
   • By “fixing the smallest-numbered unmatched vertex,” we prevent counting the same pairing multiple times.

5. Answer = \(S + dp[\text{full\_mask}]\) (total edge weight sum + minimum matching cost)

Complexity

• Time complexity: \(O(K \cdot (N + M) \log N + 2^K \cdot K)\)
  • \(K\) is the number of odd-degree vertices (\(K \leq 16\))
  • Running Dijkstra’s algorithm \(K\) times: \(O(K \cdot (N + M) \log N)\)
  • Bitmask DP: \(O(2^K \cdot K)\)
• Space complexity: \(O(N + M + 2^K)\)

Implementation Notes

• Only odd-degree vertices need to be Dijkstra sources. Since there are at most 16, this is sufficiently fast.

• The technique of “fixing the smallest-numbered unmatched vertex” in the bitmask DP ensures each pairing is considered exactly once, keeping the number of states to \(O(2^K)\).

• Although there may be multiple edges between the same pair of vertices, simply adding them to the adjacency list for Dijkstra’s algorithm works correctly. The shortest path computation automatically selects the shortest edge.

• Using a sufficiently large integer instead of float('inf') can improve performance, but with \(K \leq 16\) in this problem, it runs within the time limit without issues.

  Source Code

import heapq
import sys
from itertools import combinations

def main():
    input_data = sys.stdin.buffer.read().split()
    idx = 0
    N = int(input_data[idx]); idx += 1
    M = int(input_data[idx]); idx += 1
    
    # Build adjacency list for Dijkstra
    adj = [[] for _ in range(N + 1)]
    total_weight = 0
    degree = [0] * (N + 1)
    
    for i in range(M):
        u = int(input_data[idx]); idx += 1
        v = int(input_data[idx]); idx += 1
        w = int(input_data[idx]); idx += 1
        adj[u].append((v, w))
        adj[v].append((u, w))
        degree[u] += 1
        degree[v] += 1
        total_weight += w
    
    # Find odd-degree vertices
    odd_vertices = [v for v in range(1, N + 1) if degree[v] % 2 == 1]
    num_odd = len(odd_vertices)
    
    # If no odd vertices, the graph is Eulerian, answer is sum of all weights
    if num_odd == 0:
        print(total_weight)
        return
    
    # Compute shortest paths between all odd-degree vertices using Dijkstra
    # num_odd <= 16
    odd_set = set(odd_vertices)
    odd_index = {v: i for i, v in enumerate(odd_vertices)}
    
    # dist[i][j] = shortest path between odd_vertices[i] and odd_vertices[j]
    dist_matrix = [[0] * num_odd for _ in range(num_odd)]
    
    for i, src in enumerate(odd_vertices):
        # Dijkstra from src
        dist = [float('inf')] * (N + 1)
        dist[src] = 0
        pq = [(0, src)]
        while pq:
            d, u = heapq.heappop(pq)
            if d > dist[u]:
                continue
            for v, w in adj[u]:
                nd = d + w
                if nd < dist[v]:
                    dist[v] = nd
                    heapq.heappush(pq, (nd, v))
        
        for j, target in enumerate(odd_vertices):
            dist_matrix[i][j] = dist[target]
    
    # Minimum weight perfect matching on odd vertices using bitmask DP
    # dp[mask] = minimum cost to match all vertices in mask
    full_mask = (1 << num_odd) - 1
    dp = [float('inf')] * (1 << num_odd)
    dp[0] = 0
    
    for mask in range(full_mask + 1):
        if dp[mask] == float('inf'):
            continue
        # Find the lowest unset bit (first unmatched vertex)
        first = -1
        for b in range(num_odd):
            if not (mask & (1 << b)):
                first = b
                break
        if first == -1:
            continue
        # Try pairing first with every other unmatched vertex
        for second in range(first + 1, num_odd):
            if not (mask & (1 << second)):
                new_mask = mask | (1 << first) | (1 << second)
                cost = dp[mask] + dist_matrix[first][second]
                if cost < dp[new_mask]:
                    dp[new_mask] = cost
    
    # Answer is total weight of all edges + minimum matching cost
    print(total_weight + dp[full_mask])

if __name__ == '__main__':
    main()

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