/**
* 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. This may look elegant, but the problem is that finding connectivity
* with edge removal is difficult (O(n^4), only passing the first 2 subtasks...).
*
* After thinking for a while, I modified my solution above. First, let's consider
* all the potential edges incident to (n-1): {n-2,n-1}, {n-3,n-1}, ..., {0,n-1}. It
* only takes one of them to make node (n-1) connected to the other nodes, so we can
* simply include the last potential edge asked by the player (which is incident to
* n-1). In this way, n-1 is certainly linked to another node with a smaller index.
* Moreover, the remaining n-1 nodes' connectivity isn't affected by node n-1, so
* the same goes for the remaining n-1 nodes. Nice~
*
* Time Complexity: O(n^2)
* Implementation 1
*/
#include <bits/stdc++.h>
#include "game.h"
typedef std::vector<int> vec;
vec left;
void initialize(int n) {
left.resize(n);
for (int k = 0; k < n; k++)
left[k] = k;
}
int hasEdge(int u, int v) {
if (u > v)
std::swap(u, v);
left[v]--;
return left[v] == 0;
}
