Overview

This problem involves finding the minimum cost to depart from office \(S\), visit all \(K\) delivery destinations (up to 15), and return to \(S\) on a directed graph. It is a variation of the Traveling Salesman Problem (TSP), solved by combining precomputation of all-pairs shortest distances with bitmask DP.

Analysis

Key Observations

1. Visit order is free: Since the \(K\) delivery destinations can be visited in any order, we need to find the optimal visit order. The number of permutations of \(K\) items is \(K!\), and for \(K = 15\) this is approximately \(1.3 \times 10^{12}\), making full enumeration infeasible.

2. No need to worry about passing through intermediate points: When traveling from one delivery destination to another, if you happen to pass through another delivery destination along the way, it can be treated as “visited.” However, as long as we travel via shortest paths, what matters is only “which delivery destinations have been visited” and “which point we are currently at.”

3. Few points of interest: Although the graph can have up to 300 vertices, the only points where we need to distinguish “where we are” are \(S\) and the \(K\) delivery destinations — a total of \(K+1\) points (at most 16). All other points are merely intermediate nodes, so by precomputing all-pairs shortest distances, we can solve the problem using only the travel costs between these few “key points.”

Issues with a Naive Approach

• Trying all permutations: \(O(K!)\), computationally infeasible for \(K=15\)
• Running Dijkstra each time: possible, but computing everything at once with Floyd-Warshall is easier

Algorithm

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

Since \(N \leq 300\), we compute the shortest distance \(\mathrm{dist}[i][j]\) between all pairs of vertices in \(O(N^3)\) using the Floyd-Warshall algorithm.

Step 2: Distance Matrix for Key Points

Let the key points be \(\{S, T_1, T_2, \ldots, T_K\}\), a total of \(K+1\) points. We extract the shortest distances between them as \(d[i][j] = \mathrm{dist}[\text{nodes}[i]][\text{nodes}[j]]\) (where \(\text{nodes}[0] = S\), \(\text{nodes}[j] = T_j\)).

Step 3: Bitmask DP (Traveling Salesman Problem)

• State: \(\mathrm{dp}[\text{mask}][i]\) = minimum cost when the set of visited delivery destinations is \(\text{mask}\) (\(K\) bits) and we are currently at key point \(i\)
• Initial state: \(\mathrm{dp}[0][0] = 0\) (at \(S\), no deliveries made yet)
• Transition: Move from current position \(u\) to key point \(v\) with shortest distance \(d[u][v]\). If \(v\) is a delivery destination, set the corresponding bit.
• Answer: \(\min_i \left(\mathrm{dp}[\text{full\_mask}][i] + d[i][0]\right)\) (return to \(S\) after visiting all delivery destinations)

As a concrete example, with \(K=2\) and delivery destinations \(T_1, T_2\): - \(\text{mask} = 00\): neither visited - \(\text{mask} = 01\): only \(T_1\) visited - \(\text{mask} = 11\): both visited → return to \(S\) to finish

The number of states is \(2^K \times (K+1)\), and each state has \(K+1\) possible transitions, so the DP part runs in \(O(2^K \cdot K^2)\).

Complexity

• Time complexity: \(O(N^3 + 2^K \cdot K^2)\)
  • Floyd-Warshall: \(O(N^3) = O(300^3) = O(2.7 \times 10^7)\)
  • Bitmask DP: \(O(2^{15} \times 16^2) \approx O(8.4 \times 10^6)\)
• Space complexity: \(O(N^2 + 2^K \cdot K)\)
  • Distance matrix: \(O(N^2)\)
  • DP table: \(O(2^K \cdot K)\)

Implementation Notes

• Floyd-Warshall initialization: Distance to self is \(0\), edges that don’t exist are initialized to \(\infty\). There are no multi-edges, but take the minimum just in case.

• Handling INF: In Python, using float('inf') avoids worrying about integer arithmetic errors.

• DP transition order: Processing \(\text{mask}\) in increasing order ensures correctness, since transitions only set additional bits. However, the transition returning to \(S\) (which doesn’t increase bits) exists, so the final answer is computed after the DP is complete by adding the return cost to \(S\) from all states.

• Returning to \(S\): We need to check not only dp[full_mask][0] but also the minimum of dp[full_mask][i] + d[i][0] (since after visiting all delivery destinations, there may still be a remaining trip back to \(S\)).

  Source Code

import sys
import heapq

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
    
    INF = float('inf')
    
    # Floyd-Warshall for all-pairs shortest paths
    dist = [[INF] * (N + 1) for _ in range(N + 1)]
    for i in range(N + 1):
        dist[i][i] = 0
    
    for _ in range(M):
        u = int(input_data[idx]); idx += 1
        v = int(input_data[idx]); idx += 1
        w = int(input_data[idx]); idx += 1
        if w < dist[u][v]:
            dist[u][v] = w
    
    S = int(input_data[idx]); idx += 1
    K = int(input_data[idx]); idx += 1
    
    T = []
    for _ in range(K):
        T.append(int(input_data[idx])); idx += 1
    
    # Floyd-Warshall
    for k in range(1, N + 1):
        for i in range(1, N + 1):
            if dist[i][k] == INF:
                continue
            for j in range(1, N + 1):
                if dist[k][j] == INF:
                    continue
                nd = dist[i][k] + dist[k][j]
                if nd < dist[i][j]:
                    dist[i][j] = nd
    
    # Key nodes: S and T[0..K-1]
    # Index 0 = S, Index 1..K = T[0..K-1]
    nodes = [S] + T
    num_nodes = K + 1
    
    # Build distance matrix between key nodes
    d = [[INF] * num_nodes for _ in range(num_nodes)]
    for i in range(num_nodes):
        for j in range(num_nodes):
            d[i][j] = dist[nodes[i]][nodes[j]]
    
    # TSP with bitmask DP
    # State: (current node index in key nodes, visited bitmask of delivery points)
    # Delivery points are bits 0..K-1 corresponding to T[0..K-1]
    # We start at node index 0 (which is S), need to visit all K delivery points, return to S.
    
    full_mask = (1 << K) - 1
    
    # dp[mask][i] = minimum cost to have visited the delivery points in mask, currently at key node i
    dp = [[INF] * num_nodes for _ in range(1 << K)]
    
    # Start at S (index 0), no deliveries yet
    dp[0][0] = 0
    
    for mask in range(1 << K):
        for u in range(num_nodes):
            if dp[mask][u] == INF:
                continue
            cost_u = dp[mask][u]
            # Try going to each key node
            for v in range(num_nodes):
                if d[u][v] == INF:
                    continue
                new_cost = cost_u + d[u][v]
                if v == 0:
                    # Going back to S, no new delivery
                    new_mask = mask
                else:
                    # v corresponds to delivery point v-1
                    new_mask = mask | (1 << (v - 1))
                if new_cost < dp[new_mask][v]:
                    dp[new_mask][v] = new_cost
    
    # Answer: visit all deliveries and return to S
    ans = dp[full_mask][0]
    
    # But we also need to consider: after visiting all, we go back to S
    # dp[full_mask][0] is already the case where we're at S with all visited
    # But we might reach full_mask at some node i != 0, then go back to S
    for i in range(num_nodes):
        if dp[full_mask][i] != INF and d[i][0] != INF:
            val = dp[full_mask][i] + d[i][0]
            if val < ans:
                ans = val
    
    if ans == INF:
        print(-1)
    else:
        print(ans)

main()

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This editorial was generated by claude4.6opus-thinking.