Overview

This problem asks for the shortest time to start from town 1, visit all towns at least once, and return to town 1. It is a variant of the so-called “Traveling Salesman Problem (TSP),” but with the distinctive feature that you may pass through the same town or road multiple times.

Analysis

1. Converting “at least once” to “exactly once”

The standard Traveling Salesman Problem has the constraint of “visiting each city exactly once,” but this problem allows passing through the same town multiple times. However, if we precompute the shortest distances between all pairs of towns, we can treat “moving from town A to town B via the shortest path” as a “single move,” reducing this to a standard Traveling Salesman Problem.

For example, even if you pass through town 2 on the way from town 1 to town 3, you can think of the movement “town 1 → town 2 → town 3” as a single consolidated move, effectively reducing the problem to “visiting all towns via the shortest route.”

2. Focusing on the constraints

Notice that the number of towns \(N\) is at most \(15\), which is very small. Trying all permutations would give \(15! \approx 1.3 \times 10^{12}\), which is too slow. However, using bitmask DP (dynamic programming), we can solve it in \(O(N^2 2^N)\) time. Since \(15^2 \times 2^{15} \approx 7.3 \times 10^6\), this comfortably fits within the time limit.

Algorithm

Step 1: All-pairs shortest path computation (Floyd-Warshall algorithm)

First, from the given road information, compute the shortest travel time between any two towns \((i, j)\). This can be computed in \(O(N^3)\) using the Floyd-Warshall algorithm. If no path exists from one town to another, the distance is treated as infinity (INF).

Step 2: Bitmask DP for computing the tour path

Next, define the following DP table: dp[mask][v]: The minimum travel time when the set of towns visited so far is mask and the current town is v

• mask is a bit representation (integer) where each bit is 1 if the corresponding town has been visited, and 0 otherwise.
• Initial state: dp[1][0] = 0 (at town 1 (index 0), with only town 1 visited)
• Transition: From the current town u, choose the next town v to visit, and update dp[mask | (1 << v)][v].

Step 3: Returning to town 1 and the final answer

After computing up to the state dp[(1 << N) - 1][u] (the state where all bits are 1, meaning all towns have been visited), add the travel time from the current town u back to the starting point, town 1. The minimum among these values is the answer. If some town is unreachable, or if it is impossible to visit all towns and return, output -1.

Complexity

• Time complexity: \(O(N^3 + N^2 2^N)\)
  • \(O(N^3)\) for the Floyd-Warshall algorithm
  • \(O(N^2 2^N)\) for the bitmask DP transitions.
• Space complexity: \(O(N^2 + N 2^N)\)
  • \(O(N^2)\) for the shortest distance matrix
  • \(O(N 2^N)\) for the DP table.

Implementation Tips

• Index adjustment: The problem uses towns numbered \(1 \sim N\), but handling them as \(0 \sim N-1\) in code makes bit operations easier.

• Infinity initialization: Initialize distances and DP table values with a sufficiently large value (such as 10^{15}).

• Reachability check: Before attempting to visit all towns, it is efficient to check whether there are towns unreachable from town 1 and immediately output -1 if so.

  Source Code

import sys

# The problem asks for the minimum travel time to visit all N towns at least once,
# starting from town 1 and returning to town 1.
# This is a variation of the Traveling Salesperson Problem (TSP). 
# Since we can visit towns and roads multiple times, we can simplify the problem 
# by first calculating the shortest paths between all pairs of towns.
# Then, we solve the standard TSP on the resulting complete graph.

def solve():
    # Read all input from standard input and split into a list of strings
    try:
        input_data = sys.stdin.read().split()
    except EOFError:
        return
    
    if not input_data:
        return
    
    # N is the number of towns, M is the number of roads
    N = int(input_data[0])
    M = int(input_data[1])
    
    # Use a large integer for infinity. The maximum possible travel time is 
    # well within this limit (max N * max W = 15 * 10^6).
    INF = 10**18
    
    # dist[i][j] will store the shortest path distance between town i and town j.
    # Initialize with INF, and 0 for the distance from a town to itself.
    dist = [[INF] * N for _ in range(N)]
    for i in range(N):
        dist[i][i] = 0
        
    ptr = 2
    for _ in range(M):
        if ptr + 2 >= len(input_data):
            break
        u = int(input_data[ptr]) - 1
        v = int(input_data[ptr+1]) - 1
        w = int(input_data[ptr+2])
        # If multiple roads exist between the same two towns, keep only the shortest one.
        if w < dist[u][v]:
            dist[u][v] = w
            dist[v][u] = w
        ptr += 3
        
    # Floyd-Warshall algorithm to compute all-pairs shortest paths.
    # This allows us to treat the problem as a standard TSP on a complete graph.
    for k in range(N):
        dk = dist[k]
        for i in range(N):
            di = dist[i]
            dik = di[k]
            if dik >= INF:
                continue
            for j in range(N):
                new_dist = dik + dk[j]
                if new_dist < di[j]:
                    di[j] = new_dist
    
    # If any town is not reachable from town 1 (index 0), it is impossible to visit all towns.
    for i in range(N):
        if dist[0][i] >= INF:
            print("-1")
            return

    # DP for TSP:
    # dp[mask][u] = minimum distance to have visited the set of towns represented by 'mask',
    # with the current location being town 'u'.
    # 'mask' is a bitmask where the i-th bit is 1 if town i+1 has been visited.
    num_states = 1 << N
    dp = [[INF] * N for _ in range(num_states)]
    
    # Starting point: Town 1 (index 0), distance is 0.
    dp[1][0] = 0
    
    # Iterate through all bitmasks representing subsets of visited towns.
    for mask in range(1, num_states):
        dp_mask = dp[mask]
        
        # Optimization: Pre-calculate which towns are not yet in the current mask.
        not_in_mask = [v for v in range(N) if not (mask & (1 << v))]
        
        # For each town 'u' that could be the current location:
        for u in range(N):
            curr_dist = dp_mask[u]
            if curr_dist >= INF:
                continue
            
            # Try moving to each town 'v' that has not been visited yet.
            dist_u = dist[u]
            for v in not_in_mask:
                new_mask = mask | (1 << v)
                new_dist = curr_dist + dist_u[v]
                if dp[new_mask][v] > new_dist:
                    dp[new_mask][v] = new_dist
                        
    # The final answer is the minimum distance to visit all towns (full_mask)
    # and then return to the starting town (town 1, index 0).
    ans = INF
    full_mask = num_states - 1
    dp_full = dp[full_mask]
    dist_0 = dist[0]
    for u in range(N):
        if dp_full[u] < INF:
            # Add the shortest path distance from the last town u back to town 1.
            ans = min(ans, dp_full[u] + dist_0[u])
            
    # If ans is still INF, it means it was impossible to visit all towns and return.
    if ans >= INF:
        print("-1")
    else:
        print(ans)

if __name__ == '__main__':
    solve()

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This editorial was generated by gemini-3-flash-thinking.