/**
* A pretty straight-forward problem. The problem asks us to "confuse" the player by
* only letting him/her know the graph's connectivity at the very last moment.
*
* In order to be "between" connected and disconnected, it makes sense to have a
* small (~O(n)) number of edges. Moreover, whether each edge is present shall be
* determined by what edges are actually asked by the player. Here's my solution:
* First imagine that all edges have the state "undecided". At the beginning,
* all edges are "by default" present (we'll find out the actual ans when
* the player asks for it). When the player asks for a new edge {u,v}, we
* see whether not including {u,v} makes the graph disconnected, assuming
* that all "undecided" edges are included.
* If it does, we must include {u,v}; otherwise, we exclude it from the
* graph since there're other undecided edges that preserve connectivity.
* Finally, we flag the edge as "decided".
* One can prove by induction that nC2 queries are required before we can determine
* connectivity.
*
* Time Complexity: O(n^2)
* Implementation 1 (connectivity)
*/
#include <bits/stdc++.h>
#include "game.h"
typedef std::vector<int> vec;
int n;
std::vector<vec> graph; // -1: undecided, 0: not included, 1: present
void initialize(int _n) {
n = _n;
graph.assign(n, vec(n, -1));
}
bool connected() {
std::vector<bool> visited(n, false);
visited[0] = true;
std::queue<int> bfs_queue;
bfs_queue.emplace(0);
int non_visited = n;
while (!bfs_queue.empty()) {
int t = bfs_queue.front();
bfs_queue.pop();
non_visited--;
for (int neighb = 0; neighb < n; neighb++) {
if (!visited[neighb] && graph[neighb][t] != 0) {
visited[neighb] = true;
bfs_queue.emplace(neighb);
}
}
}
return non_visited == 0;
}
int hasEdge(int u, int v) {
if (graph[u][v] == -1) {
graph[u][v] = graph[v][u] = 0;
if (!connected())
graph[u][v] = graph[v][u] = 1;
}
return graph[u][v];
}
