Problem Overview

Given a graph with (n) intersections ((n \le 10^3)) and (m) bidirectional roads ((m \le 10^4)), a person named T moves first along a predetermined path (c_1, c_2, \ldots, c_g). Each road has a travel time (f[u][v]).

When T traverses a road, that road becomes blocked for the entire duration of T's crossing. Luka starts from intersection (a) at time (k) and needs to reach intersection (b).

If Luka arrives at a road exact when its blocked, he must wait until T finishes crossing. This effectively doubles the crossing time for blocked roads.

Solution Approach

The constraints suggest using an adjacency matrix since (n^2 = 10^6) easily fits in memory. The core idea is to precompute the blocking times for all roads, then run Dijkstra's algorithm with modified edge weights.

Precomputing Road Block Times

We simulate T's journey and record when each road becomes blocked:

int currentTime = 0;
for (int i = 1; i < pathLen; ++i) {
    blockTime[c[i]][c[i + 1]] = blockTime[c[i + 1]][c[i]] = currentTime;
    currentTime += travelTime[c[i]][c[i + 1]];
}

The blockTime[u][v] stores the exact moment when the road from (u) to (v) becomes unavailable.

Modified Edge Traversal

When traversing edge (u \to v) from a node reached at time dist[u]:

int calculateCrossingTime(int u, int v, int arrivalTime) {
    int blockStart = blockTime[u][v];
    int crossingDuration = travelTime[u][v];
    
    if (arrivalTime >= blockStart && arrivalTime <= blockStart + crossingDuration - 1) {
        // Road is blocked - Luka arrives during T's crossing
        return blockStart + 2 * crossingDuration;
    }
    return arrivalTime + crossingDuration;
}

The key insight: if Luka arrives during the blocked period, his arrival time effectively becomes blockStart + 2 * crossingDuration (waiting for T to finish plus his own crossing).

Dijkstra's Algorithm Implementation

#include <bits/stdc++.h>
#define F(i, a, b) for(int i = a; i <= b; ++i)
using namespace std;

const int MAXN = 1005;
const int INF = 1e9;

int nodeCount, roadCount, startNode, targetNode, startTime, pathLen;
int travelTime[MAXN][MAXN];
int blockTime[MAXN][MAXN];
int minDist[MAXN];
bool visited[MAXN];
int traversalOrder[MAXN];

struct NodeState {
    int vertex;
    int time;
    bool operator<(const NodeState& other) const {
        return time > other.time;
    }
};

int getCrossingTime(int from, int to, int arrivalTime) {
    int duration = travelTime[from][to];
    int blockStart = blockTime[from][to];
    
    if (arrivalTime >= blockStart && arrivalTime < blockStart + duration) {
        return blockStart + 2 * duration;
    }
    return arrivalTime + duration;
}

void shortestPath() {
    fill(minDist, minDist + MAXN, INF);
    minDist[startNode] = startTime;
    
    priority_queue<NodeState> pq;
    pq.push({startNode, startTime});
    
    while (!pq.empty()) {
        NodeState cur = pq.top();
        pq.pop();
        
        if (visited[cur.vertex]) continue;
        visited[cur.vertex] = true;
        
        F(v, 1, nodeCount) {
            if (travelTime[cur.vertex][v] > 0) {
                int arrival = getCrossingTime(cur.vertex, v, minDist[cur.vertex]);
                if (minDist[v] > arrival) {
                    minDist[v] = arrival;
                    pq.push({v, arrival});
                }
            }
        }
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    cin >> nodeCount >> roadCount >> startNode >> targetNode >> startTime >> pathLen;
    
    F(i, 1, pathLen) {
        cin >> traversalOrder[i];
    }
    
    F(i, 1, roadCount) {
        int u, v, w;
        cin >> u >> v >> w;
        travelTime[u][v] = travelTime[v][u] = w;
    }
    
    int elapsed = 0;
    F(i, 1, pathLen - 1) {
        int from = traversalOrder[i];
        int to = traversalOrder[i + 1];
        blockTime[from][to] = blockTime[to][from] = elapsed;
        elapsed += travelTime[from][to];
    }
    
    shortestPath();
    cout << (minDist[targetNode] - startTime) << "\n";
    
    return 0;
}

Algorithm Analysis

• Time Complexity: (O(n^2)) for Dijkstra with adjacency matrix, plus (O(n)) for preprocessing blocking times. With (n \le 1000), this is well within limits.
• Space Complexity: (O(n^2)) for storing the adjacency matrix and block times.

Key Observations

1. T's path blocks roads chronologically based on traversal order
2. When arriving during a block, Luka waits for T to finish, resulting in (2 \times f[u][v]) total crossing time
3. The problem transforms into a standard shortest path with dynamic edge weights

This approach efficiently handles the time-dependent road restrictions while maintaining accurate waiting time calculations.