Minimizing Dark Travel

Analysis

We have:

• A starting point $A = (a_x, a_y)$
• A destination point $B = (b_x, b_y)$
• $n$ streetlights, each streetlight $i$ has center $(x_i, y_i)$ and radius $r_i$

Alice travels at speed 1 unit/second. We want to minimize the total time spent outside any illuminated region. Equivalently, if we think of “time in darkness” as “distance traveled outside the union of circles,” we want to find a path from $A$ to $B$ that minimizes the length of its portion outside all circles.

A common competitive-programming approach to problems like this is to build a (weighted) graph where:

1. Nodes:

   • One node for the starting point $A$ (call it node 0).
   • One node for the destination point $B$ (call it node n+1).
   • One node for each of the $n$ streetlights (call them nodes 1 through n).

2. Edges & Weights:

   • Between each pair of streetlight nodes $i$ and $j$:
     • Compute the distance $d$ between their centers.
     • The “gap” you must be in darkness while moving from the coverage of streetlight $i$ to the coverage of streetlight $j$ is: $$\text{cost}(i, j) \;=\; \max\bigl(0,\; d - (r_i + r_j)\bigr).$$
     • If the two streetlights overlap or just touch ($d \le r_i + r_j$), then the cost is $0$, because you can go from one illuminated region to the other without going dark.
   • Between $A$ (node 0) and streetlight $i$ (node i): $$\text{cost}(0, i) \;=\; \max\bigl(0,\; \text{dist}(A, \text{center}_i) - r_i\bigr).$$
     • If $A$ is already inside or on the boundary of streetlight $i$, that cost is $0$.
   • Similarly, between $B$ (node n+1) and streetlight $i$ (node i): $$\text{cost}(i, n+1) \;=\; \max\bigl(0,\; \text{dist}(B, \text{center}_i) - r_i\bigr).$$
   • Finally, we must consider a direct edge from $A$ to $B$. In many formulations, you can either:
     • Add a direct edge cost from 0 to n+1 as $\text{dist}(A, B)$. (Because you could technically ignore all lights.)
     • Or just rely on possibly traveling through lights for cheaper “darkness” cost.

Once this graph is built (with $(n+2)$ nodes and up to $(n+2)(n+1)/2$ edges), run a shortest path algorithm (Dijkstra) from node 0 to node n+1. The result gives you the minimum total darkness distance. Since Alice moves at speed 1 unit/s, that distance is also the time in darkness. Finally, we take the floor of that value.

cpp
#include <bits/stdc++.h>
using namespace std;

// A simple function to compute Euclidean distance.
double dist(double x1, double y1, double x2, double y2) {
    double dx = x1 - x2;
    double dy = y1 - y2;
    return sqrt(dx*dx + dy*dy);
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    // Read the coordinates of A (start) and B (destination)
    double ax, ay;
    double bx, by;
    cin >> ax >> ay >> bx >> by;

    // Read the number of streetlights
    int n;
    cin >> n;

    // Read streetlights: each has (x, y, r)
    vector<double> lx(n), ly(n), lr(n);
    for(int i = 0; i < n; i++){
        cin >> lx[i] >> ly[i] >> lr[i];
    }

    // We will create a graph with (n+2) nodes:
    //   node 0 = A
    //   node n+1 = B
    //   nodes [1..n] = streetlights
    // We'll store adjacency in a vector of vectors (or adjacency list).
    // Each entry: (neighbor, cost).
    vector<vector<pair<int,double>>> adj(n+2);

    // Helper lambda to add edges in both directions.
    auto add_edge = [&](int u, int v, double w){
        // For an undirected setup, add both ways:
        adj[u].push_back({v, w});
        adj[v].push_back({u, w});
    };

    // 1) Add edges between each pair of streetlights
    for(int i = 0; i < n; i++){
        for(int j = i+1; j < n; j++){
            double d = dist(lx[i], ly[i], lx[j], ly[j]);
            double cost = max(0.0, d - (lr[i] + lr[j]));
            add_edge(i+1, j+1, cost);
        }
    }

    // 2) Add edges between A (node 0) and each streetlight (node i+1)
    //    cost = max(0, distance(A, center_i) - radius_i)
    for(int i = 0; i < n; i++){
        double d = dist(ax, ay, lx[i], ly[i]);
        double cost = max(0.0, d - lr[i]);
        add_edge(0, i+1, cost);
    }

    // 3) Add edges between B (node n+1) and each streetlight (node i+1)
    for(int i = 0; i < n; i++){
        double d = dist(bx, by, lx[i], ly[i]);
        double cost = max(0.0, d - lr[i]);
        add_edge(n+1, i+1, cost);
    }

    // 4) Also consider a direct edge from A to B (in case it's cheaper to go outside all lights).
    {
        double d = dist(ax, ay, bx, by);
        add_edge(0, n+1, d);
    }

    // Now we run Dijkstra from node 0 to node n+1.
    // distance array
    vector<double> minDark(n+2, numeric_limits<double>::infinity());
    minDark[0] = 0.0;

    // priority queue for Dijkstra: (darkness so far, node)
    priority_queue<pair<double,int>, vector<pair<double,int>>, greater<pair<double,int>>> pq;
    pq.push({0.0, 0});

    while(!pq.empty()){
        auto [currDark, u] = pq.top();
        pq.pop();
        if(currDark > minDark[u]) {
            continue;
        }
        if(u == n+1) {
            // We can stop if we want the distance to B immediately
            break;
        }
        // Relax edges
        for(auto &edge : adj[u]){
            int v = edge.first;
            double w = edge.second;
            double newDark = currDark + w;
            if(newDark < minDark[v]){
                minDark[v] = newDark;
                pq.push({newDark, v});
            }
        }
    }

    // The answer is minDark[n+1], the minimal "darkness distance" from A to B
    // Then we floor it as requested.
    double answer = floor(minDark[n+1]);
    cout << (long long)answer << "\n";
    return 0;
}

Explanation

1. Nodes:

   • Node 0 represents $A$.
   • Nodes 1 through $n$ represent the $n$ streetlights.
   • Node $n+1$ represents $B$.

2. Edges:

   • Between every pair of streetlights $i, j$: The cost is $\max\bigl(0, \text{dist}(center_i, center_j) - (r_i + r_j)\bigr)$.
     • If the circles overlap, $\text{dist}(center_i, center_j) \le r_i + r_j$, the cost is 0.
   • Between $A$ and streetlight $i$:
     $\max\bigl(0, \text{dist}(A, center_i) - r_i\bigr)$.
     • If $A$ is inside or on streetlight $i$, cost is 0.
   • Between $B$ and streetlight $i$:
     $\max\bigl(0, \text{dist}(B, center_i) - r_i\bigr)$.
   • Also an edge between $A$ and $B$, with cost = $\text{dist}(A, B)$.

3. Shortest Path:
   We run Dijkstra (since edges can have 0 or positive cost) from node 0 to node $n+1$. The distance we get is the minimal “dark distance.”

4. Output:
   Finally, we take floor(...) of the best distance found, and output it.

This solves the problem of minimizing time spent in darkness (equivalently, distance traveled out of coverage), returning the floor of that minimal value.