이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
using namespace std;
#define FAST_IO ios_base::sync_with_stdio(0); cin.tie(nullptr)
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define REP(n) FOR(O, 1, (n))
#define f first
#define s second
#define pb push_back
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<pii> vii;
typedef vector<ll> vl;
const int MAXN = 100100;
const ll MOD = 1000 * 1000 * 1000;
ll par[MAXN], sz[MAXN];
int nodeId[MAXN];
int find (int a) {return par[a] = par[a]==a ? a : find(par[a]);}
bool same (int a, int b) {return find(a) == find(b);}
void unite (int a, int b) {
a = find(a), b = find(b);
if (a == b) return;
par[b] = a;
sz[a] += sz[b];
}
int x[MAXN], y[MAXN];
ll n;
ll ans = 0;
unordered_set<int> adj[MAXN];
map<pair<ll, ll>, int> toId;
ll dp[MAXN];
void dfs (int v, int p) {
for (int u : adj[v]) {
if (u == p) continue;
dfs(u, v);
dp[v] += dp[u];
}
ans += dp[v]*(n-dp[v]) % MOD;
ans %= MOD;
}
int cnt = 0;
int DistanceSum(int N, int *X, int *Y) {
n = N;
FOR(i, 0, n-1) x[i] = X[i];
FOR(i, 0, n-1) y[i] = Y[i];
FOR(i, 0, n-1) toId[{x[i], y[i]}] = i+1;
/// first tree
FOR(i, 0, n-1) {
par[i] = i;
sz[i] = 1;
}
FOR(i, 0, n-1) {
int id = toId[{x[i], y[i]-1}];
if (id > 0) {
id--;
unite(i, id);
}
}
FOR(i, 0, n-1) {
if (find(i) != i) continue;
nodeId[i] = ++cnt;
dp[cnt] = sz[i];
}
FOR(i, 0, n-1) {
int id = toId[{x[i]+1, y[i]}];
if (id > 0) {
id--;
int u = nodeId[find(i)];
int v = nodeId[find(id)];
adj[u].insert(v);
adj[v].insert(u);
}
}
/// second tree
FOR(i, 0, n-1) {
par[i] = i;
sz[i] = 1;
}
FOR(i, 0, n-1) {
int id = toId[{x[i]-1, y[i]}];
if (id > 0) {
id--;
unite(i, id);
}
}
FOR(i, 0, n-1) {
if (find(i) != i) continue;
nodeId[i] = ++cnt;
dp[cnt] = sz[i];
}
FOR(i, 0, n-1) {
int id = toId[{x[i], y[i]+1}];
if (id > 0) {
id--;
int u = nodeId[find(i)];
int v = nodeId[find(id)];
adj[u].insert(v);
adj[v].insert(u);
}
}
dfs (1, -1);
dfs (cnt, -1);
return ans%MOD;
}
/*
15
7 5
7 6
8 3
8 4
8 5
8 6
9 3
9 4
9 5
9 6
10 3
10 4
10 5
10 6
11 5
ans: 278?
*/
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