結果

問題 No.1418 Sum of Sum of Subtree Size
ユーザー ぷらぷら
提出日時 2021-03-05 22:39:56
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 100 ms / 2,000 ms
コード長 2,836 bytes
コンパイル時間 1,454 ms
コンパイル使用メモリ 169,272 KB
実行使用メモリ 18,944 KB
最終ジャッジ日時 2024-10-07 03:25:54
合計ジャッジ時間 4,147 ms
ジャッジサーバーID
(参考情報)
judge4 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 41
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using pcc = pair<char,char>;
using pii = pair<int,int>;
using pll = pair<ll,ll>;
using pdd = pair<ld,ld>;
using tuplis = array<ll,3>;
template<class T> using prique = priority_queue<T,vector<T>,greater<T>>;
#define rep(i,n) for(ll i = 0; i < n; i++)
#define rrep(i,n) for(ll i = n-1; i >= 0; i--)
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define Sort(a) sort(all(a))
#define Rev(a) reverse(all(a))
#define Uniq(a) sort(all(a));a.erase(unique(all(a)),a.end())
const ll LINF = 1001001001001001001;
const int INF = 1001001001;
const int MOD = 1000000007;
const int MODD = 998244353;
const ld PI = 3.1415926535897932;
const ll dx[] = {0, 1, 0, -1, 1, -1, 1, -1};
const ll dy[] = {1, 0, -1, 0, 1, 1, -1, -1};
inline ll popcnt(ll a){ return __builtin_popcountll(a); }
inline ll intpow(ll a, ll b){ ll ans = 1; while(b){ if(b & 1) ans *= a; a *= a; b /= 2; } return ans; }
inline ll modpow(ll a, ll b, ll p){ ll ans = 1; while(b){ if(b & 1) (ans *= a) %= p; (a *= a) %= p; b /= 2; } return ans; }
ll gcd(ll a,ll b){ if(a%b == 0){return b; } else{ return gcd(b,a%b); } }
ll lcm(ll a,ll b){ return a/gcd(a,b)*b; }
template<class T> bool chmin(T& a, const T& b){ if(a > b){ a = b; return 1; } return 0; }
template<class T> bool chmax(T& a, const T& b){ if(a < b){ a = b; return 1; } return 0; }
template<class T, class U> bool chmin(T& a, const U& b){ if(a > T(b)){ a = b; return 1; } return 0; }
template<class T, class U> bool chmax(T& a, const U& b){ if(a < T(b)){ a = b; return 1; } return 0; }
ll N,cnt[100005],used[100005],cnt2[100005],used2[100005],cnt3[100005];
vector<int>ki[100005];
ll dfs(int now) {
ll sum = 0;
used[now] = true;
rep(i,ki[now].size()) {
if(used[ki[now][i]]) {
continue;
}
sum += dfs(ki[now][i]);
}
return cnt[now] = sum+1;
}
ll dfs2(int now) {
ll sum = 0;
used2[now] = true;
rep(i,ki[now].size()) {
if(used2[ki[now][i]]) {
continue;
}
sum += dfs2(ki[now][i]);
}
return cnt2[now] = sum+cnt[now];
}
ll ans = 0;
void dfs3(int now) {
ll sum = 0;
rep(i,ki[now].size()) {
if(cnt[ki[now][i]] > cnt[now]) {
sum += cnt3[ki[now][i]]-cnt2[now]-cnt[now];
}
else {
sum += cnt2[ki[now][i]];
}
}
cnt3[now] = sum+N;
ans += cnt3[now];
rep(i,ki[now].size()) {
if(cnt[ki[now][i]] > cnt[now]) {
continue;
}
else {
dfs3(ki[now][i]);
}
}
}
int main() {
cin >> N;
rep(i,N-1) {
int A,B;
cin >> A >> B;
A--;B--;
ki[A].push_back(B);
ki[B].push_back(A);
}
dfs(0);
dfs2(0);
dfs3(0);
cout << ans << endl;
}
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