2016 Dec Silver Problem 3 Moocast

Official Problem Statement

Moocast

Introduction

In the Moo-Cast problem, Farmer John's cows are setting up an emergency broadcast system using walkie-talkies. The goal is to determine the maximum number of cows that can be reached by a broadcast originating from a single cow, considering the walkie-talkies' transmission powers and the possibility of relaying messages. In this blog post, we'll discuss a graph-based approach to solving this problem.

Problem Analysis

The problem can be modeled as a directed graph, where each cow represents a node and an edge from cow A to cow B exists if cow A can directly transmit a message to cow B. The objective is to find the maximum number of cows that can be reached from a single starting cow, considering paths with multiple hops.

Solution Strategy

We can use Breadth-First Search (BFS) to traverse the directed graph and find the maximum number of cows that can be reached from each starting cow. Here's the step-by-step solution strategy:

• Create a directed graph with N nodes (cows), where each node has outgoing edges to the cows it can directly transmit a message to.
• For each cow, use BFS to traverse the graph and count the number of reachable cows.
• Keep track of the maximum number of cows reached by any single cow.

C++ Implementation

First, let's create a function to check if cow A can transmit a message to cow B:

bool canTransmit(const vector<int>& cowA, const vector<int>& cowB) {
    int dx = cowA[0] - cowB[0];
    int dy = cowA[1] - cowB[1];
    int distanceSquared = dx * dx + dy * dy;
    return distanceSquared <= cowA[2] * cowA[2];
}

Next, implement the BFS traversal to find the number of reachable cows:

int bfs(const vector<vector<int>>& cows, int start) {
    int count = 0;
    vector<bool> visited(cows.size(), false);
    queue<int> q;
    q.push(start);
    visited[start] = true;

    while (!q.empty()) {
        int current = q.front();
        q.pop();
        count++;

        for (int i = 0; i < cows.size(); ++i) {
            if (!visited[i] && canTransmit(cows[current], cows[i])) {
                visited[i] = true;
                q.push(i);
            }
        }
    }
    return count;
}

Finally, iterate over all cows and find the maximum number of reachable cows:

int main() {
    int n;
    cin >> n;
    vector<vector<int>> cows(n, vector<int>(3));
    for (int i = 0; i < n; ++i) {
        cin >> cows[i][0] >> cows[i][1] >> cows[i][2];
    }

    int maxCows = 0;
    for (int i = 0; i < n; ++i) {
        maxCows = max(maxCows, bfs(cows, i));
    }

    cout << maxCows << endl;
}

Related Questions

• USACO 2017 December Contest, Silver - Problem 2: Moocast

   This problem is similar to the Moo-Cast problem, but instead of determining the maximum number of cows that can be reached by a broadcast originating from a single cow, the goal is to find the minimum transmission power needed to connect all cows.

• USACO 2017 December Contest, Bronze - Problem 2: Blocked Billboard

   In this problem, you must determine if two rectangular billboards can be placed without overlapping, given the positions of the billboards and the positions of the obstacles.

• USACO 2012 November Contest, Bronze - Problem 3: Cow Hopscotch

   In this problem, cows play a game of hopscotch on a grid. You need to find the total number of different paths a cow can take to reach the other side of the grid, given certain restrictions on the movement.

• USACO 2019 January Contest, Silver - Problem 3: Icy Perimeter

   In this problem, you are given an icy grid and must find the largest connected ice block's perimeter. This problem involves finding connected components in a grid using depth-first search (DFS) or breadth-first search (BFS).

• USACO 2016 December Contest, Silver - Problem 3: Cities and States

   In this problem, you must count the number of city-state pairs such that the first two letters of the city name match the state abbreviation, but the city is not in the state. This problem involves graph connectivity and traversal.