Problem: Link, Submission: 280673288
Code
signed main() {
ios::sync_with_stdio(0), cin.tie(0);
int n; cin >> n;
vector<vector<int>> g(2001);
vector<pair<int, int>> a(n);
for (int i = 0; i < n; i++) {
int u, v;
cin >> u >> v;
a[i].first = u + 1000;
a[i].second = v + 1000;
g[u + 1000].push_back(v + 1000);
g[v + 1000].push_back(u + 1000);
}
int ans = 0;
for (int i = 0; i < n; i++) {
int x = a[i].first;
int y = a[i].second;
bool l = 0, r = 0, u = 0, d = 0;
for (auto k : g[y]) {
if (k > x) {
r = 1;
} else if (k < x) {
l = 1;
}
}
for (auto k : g[x]) {
if (k > y) {
u = 1;
} else if (k < y) {
d = 1;
}
}
if (l == 1 and r == 1 and u == 1 and d == 1) {
ans += 1;
}
}
std::cout << ans << "\n";
return 0;
}
I'm sorry but function definition is wrong: do try this:
public static void main(String args[])It is C++ bro.
just brute force bro. Link
You have to split your vector g into two separate vectors, one for horizontal, and one vertical. Given input:
your code would say that (0, 0) is a "supercentral point" because there exist smaller and bigger xs and ys than 0, but really there is no "right neighbour" of that point
Thank you for providing the testcase, now I understood my error.
Your approach is not correct. You are just checking for left , right, up and down points. But the condition says that they have to be direct left, right, up and down points.
point (x', y') is (x, y)'s right neighbor, if x' > x and y' = ypoint (x', y') is (x, y)'s left neighbor, if x' < x and y' = ypoint (x', y') is (x, y)'s lower neighbor, if x' = x and y' < ypoint (x', y') is (x, y)'s upper neighbor, if x' = x and y' > ySo, before checking
x > x 'orx < x'you have to checky == y'.Similarly you also need to check
x == x'fory > y'andy < y'to be valid.So, if you really want to use this approach, you can use two seperate g vectors gx and gy.
see my edited submission of your code here 280712862. (accepted)
just check all the points. 200 is not that much points. My submission 280695175.
Yeah, my approach was wrong. Thanks for the correction and the new solution.