Simply find the minimum distance from each room to \(A\) and \(B\) respectively using a BFS. The room with the minimum \(\max(\text{Dist to A}, \text{Dist to B})\) will be the answer.
Time: \(O(V + E)\)
Memory: \(O(V + E)\)
#include <bits/stdc++.h>
#define FOR(i, x, y) for (int i = x; i < y; i++)
using namespace std;
typedef long long ll;
vector<int> graph[10001];
int visited[10001][2];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int n, m, a, b;
cin >> n >> m >> a >> b;
FOR(i, 0, m) {
int u, v;
cin >> u >> v;
graph[u].push_back(v);
graph[v].push_back(u);
}
queue<int> q;
q.push(a);
visited[a][0] = 1;
while (q.size()) {
int curr = q.front();
q.pop();
for (int i : graph[curr]) {
if (!visited[i][0]) {
visited[i][0] = visited[curr][0] + 1;
q.push(i);
}
}
}
q.push(b);
visited[b][1] = 1;
while (q.size()) {
int curr = q.front();
q.pop();
for (int i : graph[curr]) {
if (!visited[i][1]) {
visited[i][1] = visited[curr][1] + 1;
q.push(i);
}
}
}
int ans = INT_MAX, room = -1;
FOR(i, 1, n + 1) {
if (visited[i][0] && visited[i][1]) {
if (ans > max(visited[i][0] - 1, visited[i][1] - 1)) ans = max(visited[i][0] - 1, visited[i][1] - 1), room = i;
}
}
cout << room;
return 0;
}