結果
| 問題 |
No.2115 Making Forest Easy
|
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2022-11-06 23:02:25 |
| 言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
AC
|
| 実行時間 | 376 ms / 2,000 ms |
| コード長 | 5,360 bytes |
| コンパイル時間 | 3,593 ms |
| コンパイル使用メモリ | 236,760 KB |
| 実行使用メモリ | 45,056 KB |
| 最終ジャッジ日時 | 2024-07-20 09:57:18 |
| 合計ジャッジ時間 | 18,200 ms |
|
ジャッジサーバーID (参考情報) |
judge3 / judge1 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 2 |
| other | AC * 50 |
ソースコード
#include<bits/stdc++.h>
#include<atcoder/all>
#define rep(i,b) for(int i=0;i<b;i++)
#define rrep(i,b) for(int i=b-1;i>=0;i--)
#define rep1(i,b) for(int i=1;i<b;i++)
#define repx(i,x,b) for(int i=x;i<b;i++)
#define rrepx(i,x,b) for(int i=b-1;i>=x;i--)
#define fore(i,a) for(auto i:a)
#define fore1(i,a) for(auto &i:a)
#define rng(x) (x).begin(), (x).end()
#define rrng(x) (x).rbegin(), (x).rend()
#define sz(x) ((int)(x).size())
#define pb push_back
#define fi first
#define se second
#define pcnt __builtin_popcountll
using namespace std;
using namespace atcoder;
using ll = long long;
using ld = long double;
template<typename T> using mpq = priority_queue<T, vector<T>, greater<T>>;
template<typename T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<typename T> bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
template<typename T> ll sumv(const vector<T>&a){ll res(0);for(auto&&x:a)res+=x;return res;}
bool yn(bool a) { if(a) {cout << "Yes" << endl; return true;} else {cout << "No" << endl; return false;}}
#define dame { cout << "No" << endl; return;}
#define dame1 { cout << -1 << endl; return;}
#define deb(x,y) cout << x << " " << y << endl;
#define debp(p) cout << p.fi << " " << p.se << endl;
#define out cout << ans << endl;
#define outd cout << fixed << setprecision(20) << ans << endl;
#define outm cout << ans.val() << endl;
#define outv fore(yans , ans) cout << yans << "\n";
#define outdv fore(yans , ans) cout << yans.val() << "\n";
#define showv(v) {fore(vy , v) {cout << vy << " ";} cout << endl;}
#define showv2(v) fore(vy , v) cout << vy << "\n";
#define showvm(v) {fore(vy , v) {cout << vy.val() << " ";} cout << endl;}
#define showvm2(v) fore(vy , v) cout << vy.val() << "\n";
using pll = pair<ll,ll>;using pil = pair<int,ll>;using pli = pair<ll,int>;using pii = pair<int,int>;using pdd = pair<ld,ld>;
using tp = tuple<int ,int ,int>;
using vi = vector<int>;using vd = vector<ld>;using vl = vector<ll>;using vs = vector<string>;using vb = vector<bool>;
using vpii = vector<pii>;using vpli = vector<pli>;using vpll = vector<pll>;using vpil = vector<pil>;
using vvi = vector<vector<int>>;using vvl = vector<vector<ll>>;using vvs = vector<vector<string>>;using vvb = vector<vector<bool>>;
using vvpii = vector<vector<pii>>;using vvpli = vector<vector<pli>>;using vvpll = vector<vpll>;using vvpil = vector<vpil>;
using mint = modint998244353;
using vm = vector<mint>;
using vvm = vector<vector<mint>>;
vector<int> dx={1,0,-1,0,1,1,-1,-1},dy={0,1,0,-1,1,-1,1,-1};
ll gcd(ll a, ll b) { return a?gcd(b%a,a):b;}
ll lcm(ll a, ll b) { return a/gcd(a,b)*b;}
#define yes {cout <<"Yes"<<endl;}
#define yesr { cout <<"Yes"<<endl; return;}
#define no {cout <<"No"<<endl;}
#define nor { cout <<"No"<<endl; return;}
const double eps = 1e-10;
const ll LINF = 1001002003004005006ll;
const int INF = 1001001001;
#ifdef MY_LOCAL_DEBUG
#define show(x) cerr<<#x<<" = "<<x<<endl
#define showp(p) cerr<<#p<<" = "<<p.fi<<" : "<<p.se<<endl
#define show2(x,y) cerr<<#x<<" = "<<x<<" : "<<#y<<" = "<<y<<endl
#define show3(x,y,z) cerr<<#x<<" = "<<x<<" : "<<#y<<" = "<<y<<" : "<<#z<<" = "<<z<<endl
#define show4(x,y,z,x2) cerr<<#x<<" = "<<x<<" : "<<#y<<" = "<<y<<" : "<<#z<<" = "<<z<<" : "<<#x2<<" = "<<x2<<endl
#define test(x) cout << "test" << x << endl
#else
#define show(x)
#define showp(p)
#define show2(x,y)
#define show3(x,y,z)
#define show4(x,y,z,x2)
#define test(x)
#endif
void solve(){
int n; cin>>n;
vi a(n);
rep(i,n) cin>>a[i];
vvi e(n);
rep(i,n-1){
int a,b; cin>>a>>b;
a--; b--;
e[a].pb(b);
e[b].pb(a);
}
const int mxa = 1001;
vvm dp1(n , vm(mxa)),dp2(n , vm(mxa));
vi dd(n);
auto dfs = [&](auto dfs,int x,int pre)->int{
int d = 1;
dp1[x][a[x]] = 1;
dp2[x][a[x]] = 1;
fore(y , e[x]){
if(y == pre) continue;
int k = dfs(dfs , y ,x);
vm tmp1(mxa);
vm tmp2(mxa);
swap(tmp1 , dp1[x]);
swap(tmp2 , dp2[x]);
vm sumx1(mxa),sumy1(mxa),sumx2(mxa),sumy2(mxa);
dp1[y][0] = mint(2).pow(k-1);
dp2[y][0] = 0;
rep(j,mxa){
sumx1[j] = tmp1[j];
sumy1[j] = dp1[y][j];
sumx2[j] = tmp2[j];
sumy2[j] = dp2[y][j];
if (j != 0){
sumx1[j] += sumx1[j-1];
sumy1[j] += sumy1[j-1];
sumx2[j] += sumx2[j-1];
sumy2[j] += sumy2[j-1];
}
}
rep(j,mxa){
dp1[x][j] = sumx1[j]*dp1[y][j] + tmp1[j]*sumy1[j] - tmp1[j]*dp1[y][j];
dp2[x][j] += sumx2[j]*dp1[y][j] + tmp1[j]*sumy2[j];
dp2[x][j] += sumx1[j]*dp2[y][j] + tmp2[j]*sumy1[j];
dp2[x][j] -= tmp1[j]*dp2[y][j] + tmp2[j]*dp1[y][j];
}
d += k;
}
dd[x] = d;
return d;
};
dfs(dfs,0,-1);
mint ans = 0;
rep(i,n) rep1(j,mxa){
int ik = 0;
if (i == 0) ik = 1;
ans += dp2[i][j] * j * mint(2).pow(n-1-dd[i]+ik);
}
outm;
return;
}
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
int t = 1;
//cin>>t;
rep(i,t){
solve();
}
return 0;
}