Consider a graph with $N$ vertices $1,2, \ldots, N$, where there is an undirected edge between vertex $i$ and $j$ if and only if the distance between $(X_i,Y_i)$ and $(X_j,Y_j)$ is $D$ or less.

Person $i$ is infected with the virus if and only if there is a path from vertex $1$ to vertex $i$ in the graph defined above, i.e. vertex $1$ and vertex $i$ belongs to the same connected component.

Therefore, we can obtain the correct answer by a graph-searching algorithm such as DFS (Depth-First Search), BFS (Breadth-First Search), or DSU (Disjoint Set Union).

If you use DFS or BFS, the time complexity is $O(N^2)$ because there are $N$ vertices and $O(N^2)$ edges; if you use DSU, the complexity is a bit worse, but it is still fast enough.

Sample code

#include <iostream>
#include <vector>
#include <queue>
using namespace std;
int main() {
	int n, d;
	cin >> n >> d;
	vector<int> x(n), y(n);
	for (int i = 0; i < n; i++) cin >> x[i] >> y[i];
	vector<vector<bool>> graph(n, vector<bool>(n));
	for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) if ((x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]) <= d * d) graph[i][j] = true;
	vector<bool> ans(n);
	ans[0] = true;
	queue<int> que; que.push(0);
	while (!que.empty()) {
		int q = que.front(); que.pop();
		for (int i = 0; i < n; i++) {
			if (graph[q][i] && !ans[i]) {
				ans[i] = true;
				que.push(i);
			}
		}
	}
	for (int i = 0; i < n; i++) cout << (ans[i] ? "Yes" : "No") << '\n';
}

#include <iostream>
#include <vector>
#include <queue>
using namespace std;

int main() {
	int n, d;
	cin >> n >> d;
	vector<int> x(n), y(n);
	for (int i = 0; i < n; i++) cin >> x[i] >> y[i];
	vector<vector<bool>> graph(n, vector<bool>(n));
	for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) if ((x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]) <= d * d) graph[i][j] = true;
	vector<bool> ans(n);
	ans[0] = true;
	queue<int> que; que.push(0);
	while (!que.empty()) {
		int q = que.front(); que.pop();
		for (int i = 0; i < n; i++) {
			if (graph[q][i] && !ans[i]) {
				ans[i] = true;
				que.push(i);
			}
		}
	}
	for (int i = 0; i < n; i++) cout << (ans[i] ? "Yes" : "No") << '\n';
}