217A - Ice Skating - CodeForces Solution

brute force dfs and similar dsu graphs *1200

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Python Code: 

N = 100 + 10
fa, x, y, ans = [0] * N, [0] * N, [0] * N, 0


def find(root):
    if root == fa[root]: return root

    fa[root] = find(fa[root])
    return fa[root]


def merge(a, b):
    ra, rb = find(a), find(b)
    if ra == rb: return
    fa[ra] = rb


n = int(input())
for i in range(1, n + 1):
    _x, _y = map(int, input().split())
    x[i] = _x
    y[i] = _y

for i in range(1, n + 1): fa[i] = i

for a in range(1, n + 1):
    for b in range(a + 1, n + 1):
        if x[a] == x[b] or y[a] == y[b]:
            merge(a, b)

for i in range(1, n + 1):
    if fa[i] == i: ans += 1

print(str(ans - 1))

C++ Code: 

#include <bits/stdc++.h>
using namespace std;
#define futaba ios_base::sync_with_stdio(false); cin.tie(NULL);
#define rio return 0;
#define fi first
#define se second

// Fun things are fun. //

pair<int, int> a[105];
vector<int> adj[105];
bool vis[105];

void dfs(int n) {
    vis[n] = true;
    for(auto i : adj[n]) {
        if(!vis[i]) {
            dfs(i);
        }
    }
}

int main() {
    /* freopen(".txt", "r", stdin);
    freopen(".txt", "w", stdout); */
    futaba
    int n;
    cin >> n;
    for(int i = 0; i < n; i++) cin >> a[i].fi >> a[i].se;
    for(int i = 0; i < n; i++) {
        for(int j = i + 1; j < n; j++) {
            if(a[i].fi == a[j].fi) {
                adj[i].push_back(j);
                adj[j].push_back(i);
            }
            if(a[i].se == a[j].se) {
                adj[i].push_back(j);
                adj[j].push_back(i);
            }
        }
    }
    int ans = 0;
    for(int i = 0; i < n; i++) vis[i] = false;
    for(int i = 0; i < n; i++) {
        if(!vis[i]) {
            dfs(i);
            ans++;
        }
    }
    ans--;
    cout << ans << '\n';
    rio
}


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