Overview

This is a weighted shortest path problem where we find the minimum cost (sum of road travel times + tolls) from intersection \(1\) to intersection \(N\), taking into account the tolls at intersections.

Analysis

Identifying the Essence of the Problem

In this problem, there are two types of movement costs:

1. Road travel times (edge weights)
2. Intersection tolls (vertex weights)

Standard Dijkstra’s algorithm only handles “edge weights,” but this problem also involves “vertex weights.”

Integrating Vertex Costs into Edge Costs

The key insight is that by incorporating the vertex costs (tolls) into the edge movement costs, we can use Dijkstra’s algorithm as-is.

Specifically, when traveling from intersection \(u\) through a road (with travel time \(w\)) to arrive at intersection \(v\), we consider the cost as follows:

\[\text{transition cost} = w + \text{toll}[v]\]

In other words, we treat “road travel time + destination toll” as the edge cost. The toll at the starting point (intersection \(1\)) is added as the initial cost at the beginning.

Concrete Example

For example, with 3 intersections: - Roads: \(1 \to 2\) (travel time \(5\)), \(2 \to 3\) (travel time \(3\)) - Tolls: intersection \(2\) has a toll of \(10\)

In this case, the total cost of \(1 \to 2 \to 3\) is \(\text{toll}[1] + 5 + \text{toll}[2] + 3 + \text{toll}[3] = 0 + 5 + 10 + 3 + 0 = 18\).

This is correctly computed by Dijkstra’s transitions: starting from \(\text{dist}[1] = \text{toll}[1] = 0\), we get \(\text{dist}[2] = 0 + 5 + 10 = 15\), \(\text{dist}[3] = 15 + 3 + 0 = 18\).

Algorithm

1. Store the toll for each intersection in an array \(\text{toll}\) (intersections without toll areas have \(0\)).
2. Run Dijkstra’s algorithm:
   • Initial state: \(\text{dist}[1] = \text{toll}[1]\) (toll at the starting point)
   • Transition: when going from intersection \(u\) to intersection \(v\) via an edge (weight \(w\)), update with \(\text{dist}[v] = \text{dist}[u] + w + \text{toll}[v]\)
3. \(\text{dist}[N]\) is the answer.

This yields the cost of the path that minimizes the total of all intersection tolls passed through and all road travel times.

Complexity

• Time complexity: \(O((N + M) \log N)\) (standard complexity of Dijkstra’s algorithm)
• Space complexity: \(O(N + M)\) (adjacency list and distance array)

Implementation Notes

• Fast input: Since \(N, M\) can be up to \(2 \times 10^5\), we speed up I/O by reading all input at once with sys.stdin.buffer.read().

• Early termination: In Dijkstra’s algorithm, the shortest distance to node \(N\) is finalized the first time it is extracted from the heap, so we terminate early with if u == N: break.

• Integration of vertex costs: By adding the destination node’s toll to the edge cost, we can solve the problem within the standard Dijkstra’s algorithm framework without any special handling. Note that only the toll at the starting point needs to be added separately as the initial value.

  Source Code

import heapq
import sys

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
    K = int(input_data[idx]); idx += 1
    
    adj = [[] for _ in range(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))
    
    toll = [0] * (N + 1)
    for j in range(K):
        L = int(input_data[idx]); idx += 1
        C = int(input_data[idx]); idx += 1
        toll[L] = C
    
    # Dijkstra with cost = sum of edge weights + sum of tolls at visited nodes
    # When we visit a node, we add its toll to the cost
    INF = float('inf')
    dist = [INF] * (N + 1)
    dist[1] = toll[1]
    heap = [(dist[1], 1)]
    
    while heap:
        d, u = heapq.heappop(heap)
        if d > dist[u]:
            continue
        if u == N:
            break
        for v, w in adj[u]:
            nd = d + w + toll[v]
            if nd < dist[v]:
                dist[v] = nd
                heapq.heappush(heap, (nd, v))
    
    print(dist[N])

main()

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