Overview

Starting from city 1, visit multiple cities within time limit \(T\) to satisfy mission conditions and maximize rewards. By exploiting the fact that at most 14 distinct cities appear across all missions, we solve this using a bitmask DP approach similar to the Traveling Salesman Problem (TSP).

Analysis

Key Observations

1. Few important cities: The number of distinct cities appearing in all missions is at most 14. Including city 1, there are at most 15. Only whether these cities have been visited matters for mission completion.

2. Problem with naive approach: With up to \(N \leq 10^4\) cities, exhaustively considering all visit orders is impossible. However, since only “which important cities have been visited” affects rewards, we only need to consider subsets of important cities.

3. Using shortest paths: Travel between important cities is optimal when using shortest distances regardless of intermediate routes. Therefore, we can precompute shortest distances between important cities using Dijkstra’s algorithm.

4. TSP with bitmask DP: Since there are at most 15 important cities, we can manage the visited set with a bitmask DP (state space \(2^{15} \times 15\)) to compute the minimum cost of visiting each subset.

Algorithm

1. Identify important cities: Collect all cities appearing in missions along with city 1, and assign indices (at most 15 cities).

2. Precompute shortest distances: Run Dijkstra from each important city to build the shortest distance matrix \(\text{dist}[i][j]\) between important cities.

3. Bitmask DP:

   • State: \(dp[\text{mask}][i]\) = minimum travel time when the visited set of important cities is \(\text{mask}\) and currently at city \(i\)
   • Initial state: \(dp[1 \ll \text{start}][\text{start}] = 0\) (at city 1 with cost 0)
   • Transition: Move to unvisited important city \(v\) → \(dp[\text{mask} | (1 \ll v)][v] = \min(dp[\text{mask}][u] + \text{dist}[u][v])\)

4. Aggregate minimum cost for each mask: \(\min\_cost[\text{mask}] = \min_i dp[\text{mask}][i]\)

5. Reward calculation: For each mask, compute the total reward of missions whose conditions are satisfied. Among masks with cost at most \(T\), output the one with the maximum reward.

Complexity

• Time complexity: \(O(L \cdot (N + M) \log N + 2^L \cdot L^2 + 2^L \cdot K)\)
  • \(L\) is the number of important cities (\(\leq 15\))
  • Dijkstra: \(L\) executions for \(O(L \cdot (N + M) \log N)\)
  • Bitmask DP: \(O(2^L \cdot L^2)\)
  • Reward aggregation: \(O(2^L \cdot K)\)
• Space complexity: \(O(2^L \cdot L + N)\)

Implementation Notes

• dist[u][v] must be computed with Dijkstra: Even if there is no direct edge between important cities, a shortest path through intermediate general cities may exist. You must not contract the graph to only important cities.

• Handling revisits: The DP transitions only to unvisited cities, and this is correct. This is because \(\text{dist}[u][v]\) already represents the shortest path through any intermediate cities (including already-visited ones), so routes that detour through previously visited cities are implicitly considered.

• City 1 is always visited: Since it is the starting point, the bit for city 1 is set in the initial bitmask state. Even if city 1 is included in a mission’s requirements, it is automatically satisfied.

• Handling INF: Since unreachable city pairs may exist, take care not to perform transitions when the distance remains INF.

  Source Code

import heapq
import sys
from itertools import combinations

def main():
    input_data = sys.stdin.buffer.read().split()
    idx = 0
    def rd():
        nonlocal idx
        val = int(input_data[idx]); idx += 1; return val

    N = rd(); M = rd(); K = rd(); T = rd()

    # Build adjacency list
    graph = [[] for _ in range(N + 1)]
    for _ in range(M):
        u = rd(); v = rd(); w = rd()
        graph[u].append((v, w))

    # Read missions
    missions = []
    all_cities = set()
    for _ in range(K):
        sz = rd()
        cities = []
        for __ in range(sz):
            cities.append(rd())
        p = rd()
        missions.append((cities, p))
        all_cities.update(cities)

    # Important cities: all cities appearing in missions + city 1
    all_cities.add(1)
    important = sorted(all_cities)
    num_imp = len(important)
    
    # Map important cities to indices
    city_to_idx = {}
    for i, c in enumerate(important):
        city_to_idx[c] = i

    # Compute shortest paths between all pairs of important cities using Dijkstra
    # We need dist[i][j] = shortest path from important[i] to important[j]
    INF = float('inf')
    dist = [[INF] * num_imp for _ in range(num_imp)]

    for si in range(num_imp):
        src = important[si]
        # Dijkstra from src
        d = [INF] * (N + 1)
        d[src] = 0
        pq = [(0, src)]
        while pq:
            cost, u = heapq.heappop(pq)
            if cost > d[u]:
                continue
            for v, w in graph[u]:
                if d[u] + w < d[v]:
                    d[v] = d[u] + w
                    heapq.heappush(pq, (d[v], v))
        for di in range(num_imp):
            dist[si][di] = d[important[di]]

    # Now we have important cities indexed 0..num_imp-1
    # City 1 is at index city_to_idx[1]
    start = city_to_idx[1]

    # Mission cities as bitmasks over important city indices
    # For each mission, compute the bitmask of required cities (excluding city 1 if present, 
    # since city 1 is always visited... actually city 1 IS always visited)
    # Actually, we use bitmask over all important cities. City 1 is always visited.
    
    # Important cities that are NOT city 1 (these are the ones we need to decide to visit)
    # But for TSP we consider all important cities.
    
    # Bitmask TSP: state = (current_city_index, visited_set_bitmask), value = min cost
    # visited_set_bitmask tracks which important cities have been visited
    # Start: at start, visited = {start}
    
    # num_imp <= 16 (15 mission cities + city 1)
    
    full = 1 << num_imp
    # dp[mask][i] = minimum cost to have visited exactly the important cities in mask, 
    # currently at important city i
    dp = [[INF] * num_imp for _ in range(full)]
    start_mask = 1 << start
    dp[start_mask][start] = 0

    for mask in range(full):
        for u in range(num_imp):
            if dp[mask][u] >= INF:
                continue
            if not (mask & (1 << u)):
                continue
            # Try going to another important city v
            for v in range(num_imp):
                if mask & (1 << v):
                    continue  # already visited (optimization, but revisiting is ok for path)
                new_mask = mask | (1 << v)
                new_cost = dp[mask][u] + dist[u][v]
                if new_cost < dp[new_mask][v]:
                    dp[new_mask][v] = new_cost
            # Also try going to already-visited city (for shorter path through it)
            # Actually we don't need to revisit already-visited important cities in TSP
            # because dist[u][v] already gives shortest path (possibly through visited cities)

    # For each mask, find the minimum cost to achieve it
    min_cost_for_mask = [INF] * full
    for mask in range(full):
        for u in range(num_imp):
            if dp[mask][u] < min_cost_for_mask[mask]:
                min_cost_for_mask[mask] = dp[mask][u]

    # For each mission, compute bitmask and reward
    mission_masks = []
    for cities, p in missions:
        bm = 0
        for c in cities:
            bm |= 1 << city_to_idx[c]
        mission_masks.append((bm, p))

    # Precompute reward for each visited mask
    reward_for_mask = [0] * full
    for mask in range(full):
        for bm, p in mission_masks:
            if (mask & bm) == bm:
                reward_for_mask[mask] += p

    ans = 0
    for mask in range(full):
        if not (mask & (1 << start)):
            continue
        if min_cost_for_mask[mask] <= T:
            if reward_for_mask[mask] > ans:
                ans = reward_for_mask[mask]

    print(ans)

main()

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