Overview

This problem asks for the shortest time to start from town \(1\), visit all towns at least once, and return to town \(1\). This is a variant of the Traveling Salesman Problem (TSP), and given the constraint \(N \leq 15\), it can be solved using bitmask DP.

Analysis

Key Observations

1. Since the same roads and towns can be traversed any number of times, the cost of moving from town \(u\) to town \(v\) can be considered as “the shortest distance from \(u\) to \(v\).” It is perfectly fine to pass through other towns along the way. In other words, if we first compute the shortest distances between all pairs of vertices, we can reduce the problem to one on a complete graph where we can move directly between any two towns at the shortest distance.

2. After this reduction, the problem becomes “starting from town \(1\) on a complete graph, visit all towns at least once and return to town \(1\) with the shortest path.” This is precisely the Traveling Salesman Problem.

Issue with the Naive Approach

If we try all possible visitation orders, the number of permutations is \((N-1)!\). When \(N = 15\), \((14)! \approx 8.7 \times 10^{10}\), which is far too large to handle in time.

Solution

We use bitmask DP. By managing “which set of towns have been visited” as a bitmask, we can limit the number of states to \(2^N \times N\). When \(N = 15\), \(2^{15} \times 15 = 491520\), which is sufficiently fast.

Algorithm

Step 1: All-Pairs Shortest Distances (Floyd-Warshall)

Initialize the adjacency matrix \(\text{dist}[i][j]\) and compute the shortest distances between all pairs of towns using the Floyd-Warshall algorithm.

\[\text{dist}[i][j] = \min(\text{dist}[i][j],\ \text{dist}[i][k] + \text{dist}[k][j])\]

This is updated for all intermediate vertices \(k\).

At this stage, if there exists a town unreachable from town \(0\) (= town 1), the answer is -1.

Step 2: Bitmask DP (TSP)

• State: \(\text{dp}[S][u]\) = the minimum travel time when the set of visited towns is \(S\) (bitmask) and we are currently at town \(u\)
• Initial state: \(\text{dp}[\{0\}][0] = 0\) (only town \(0\) has been visited, we are at town \(0\), cost \(0\))
• Transition: Move from town \(u\) to town \(v\)

\[\text{dp}[S \cup \{v\}][v] = \min(\text{dp}[S \cup \{v\}][v],\ \text{dp}[S][u] + \text{dist}[u][v])\]

• Answer: For the state \(S = \text{full\_mask}\) where all towns have been visited

\[\text{ans} = \min_{u} (\text{dp}[\text{full\_mask}][u] + \text{dist}[u][0])\]

In other words, it is the minimum value after adding the cost of returning to town \(0\) after visiting all towns.

Complexity

• Time complexity: \(O(N^3 + 2^N \cdot N^2)\)
  • Floyd-Warshall: \(O(N^3)\)
  • Bitmask DP: For \(O(2^N \cdot N)\) states, each with \(N\) possible transitions → \(O(2^N \cdot N^2)\)
  • When \(N = 15\), \(2^{15} \times 15^2 \approx 7.4 \times 10^6\), which is sufficiently fast
• Space complexity: \(O(2^N \cdot N)\)

Implementation Notes

• Town numbers are \(1\)-indexed in the input, but converting to \(0\)-indexed internally makes them easier to handle

• Don’t forget to initialize dist[i][i] = 0 before running Floyd-Warshall

• During DP transitions, check S & (1 << u) to verify that the current location \(u\) is included in the visited set (to prevent transitions from invalid states)

• In the final answer, don’t forget to add the return cost from the current location back to town \(0\) after visiting all towns

  Source Code

import heapq
import sys

def solve():
    input_data = sys.stdin.read().split()
    idx = 0
    N = int(input_data[idx]); idx += 1
    M = int(input_data[idx]); idx += 1
    
    INF = float('inf')
    
    # Floyd-Warshall for all-pairs shortest paths
    dist = [[INF] * N for _ in range(N)]
    for i in range(N):
        dist[i][i] = 0
    
    for _ in range(M):
        u = int(input_data[idx]) - 1; idx += 1
        v = int(input_data[idx]) - 1; idx += 1
        w = int(input_data[idx]); idx += 1
        if w < dist[u][v]:
            dist[u][v] = w
            dist[v][u] = w
    
    for k in range(N):
        for i in range(N):
            if dist[i][k] == INF:
                continue
            for j in range(N):
                if dist[k][j] == INF:
                    continue
                if dist[i][k] + dist[k][j] < dist[i][j]:
                    dist[i][j] = dist[i][k] + dist[k][j]
    
    # Check if all towns are reachable from town 0
    for i in range(N):
        if dist[0][i] == INF:
            print(-1)
            return
    
    # TSP with bitmask DP
    # dp[S][i] = minimum cost to have visited the set of towns S, currently at town i
    # S is a bitmask of visited towns
    # We start at town 0, need to visit all towns, and return to town 0
    
    full_mask = (1 << N) - 1
    dp = [[INF] * N for _ in range(1 << N)]
    dp[1 << 0][0] = 0
    
    for S in range(1 << N):
        for u in range(N):
            if dp[S][u] == INF:
                continue
            if not (S & (1 << u)):
                continue
            for v in range(N):
                if dist[u][v] == INF:
                    continue
                nS = S | (1 << v)
                cost = dp[S][u] + dist[u][v]
                if cost < dp[nS][v]:
                    dp[nS][v] = cost
    
    ans = INF
    for u in range(N):
        if dp[full_mask][u] == INF:
            continue
        if dist[u][0] == INF:
            continue
        ans = min(ans, dp[full_mask][u] + dist[u][0])
    
    if ans == INF:
        print(-1)
    else:
        print(ans)

solve()

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