結果
問題 | No.2360 Path to Integer |
ユーザー |
|
提出日時 | 2023-06-23 22:43:52 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 254 ms / 2,500 ms |
コード長 | 7,060 bytes |
コンパイル時間 | 4,192 ms |
コンパイル使用メモリ | 242,812 KB |
実行使用メモリ | 51,124 KB |
最終ジャッジ日時 | 2024-07-01 02:29:00 |
合計ジャッジ時間 | 6,922 ms |
ジャッジサーバーID (参考情報) |
judge4 / judge1 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 2 |
other | AC * 15 |
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 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_popcountllusing 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 cout2(x,y) cout << x << " " << y << endl;#define coutp(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 coutv(v) {fore(vy , v) {cout << vy << " ";} cout << endl;}#define coutv2(v) fore(vy , v) cout << vy << "\n";#define coutvm(v) {fore(vy , v) {cout << vy.val() << " ";} cout << endl;}#define coutvm2(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 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 mint = modint1000000007;//using mint = dynamic_modint<0>;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#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";#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)#define showv(v)#define showv2(v)#define showvm(v)#define showvm2(v)#endifvs a;mint ans;template<typename T>struct ReRooting {using F1 = function<T(T, T)>;using F2 = function<T()>;int n; // 頂点数vvi e,tmp_edge; // e:edge情報 親は最後に入るvector<int> d;vector<vector<T>> dp; // dp[x][i]、iはeに連動vector<vector<T>> lsum, rsum; // 左(右)からの累積和vector<T> ans2;F1 op; // マージ演算関数F2 id; // 単位元ReRooting(int n,F1 op,F2 id) : n(n),e(n),tmp_edge(n),dp(n),lsum(n),rsum(n),ans2(n),op(op),id(id),d(n){}void add_edge(int u, int v) {tmp_edge[u].pb(v);tmp_edge[v].pb(u);}T dfs1(int x, int pre) {T ret = id();int i = 0;d[x] = 1;fore(y , tmp_edge[x]){if (y == pre) continue;dp[x].pb(dfs1(y , x));d[x] += d[y];e[x].pb(y);ret = op(ret , dp[x][i]);lsum[x].pb(id());rsum[x].pb(id());i++;}// ll l = stoll(a[x]);ret++;// ans += mint(l)*ret;ret *= mint(10).pow(sz(a[x]));if (pre != -1){e[x].pb(pre);dp[x].pb(id());lsum[x].pb(id());rsum[x].pb(id());}return ret+id();}void dfs2(int x,int pre,T preval) {int m = sz(e[x]);ll l = stoll(a[x]);if (pre!=-1) ans += preval*l*d[x];if (pre != -1) dp[x][m-1] = preval;rep(i,m){lsum[x][i] = dp[x][i];if (i != 0) lsum[x][i] = op(lsum[x][i], lsum[x][i-1]);}rrep(i,m){rsum[x][i] = dp[x][i];if (i != m-1) rsum[x][i] = op(rsum[x][i], rsum[x][i+1]);}if (m >= 1) ans2[x] = lsum[x][m-1];else ans2[x] = id();rep(i,m){int y = e[x][i];if (y == pre) continue;ans += dp[x][i]*l*(n-d[y]);T nxtval = id();if (i != 0) nxtval = op(nxtval , lsum[x][i-1]);if (i != m-1) nxtval = op(nxtval , rsum[x][i+1]);nxtval++;nxtval *= mint(10).pow(sz(a[x]));dfs2(y , x , nxtval+id());}ans += mint(l)*n;return;}void solve(){dfs1(0,-1);dfs2(0,-1,id());}};// 全方位木dp// ReRooting<T> rr(n,op,id) : 初期化。nは頂点数、opはmerge演算関数、idは単位元// op,idはラムダ式で定義// add_edge(a,b) a,b間に辺を引く// solve() dp実行、ansに答えが入る。// dfs内は適宜変更することmint op(mint a,mint b){return a+b;}mint id(){return 0;}void solve(){int n; cin>>n;a.resize(n);ans = 0;rep(i,n) cin>>a[i];// auto op = [](mint a,mint b)->mint{// return a+b;// };// auto id = []()->mint{// return mint(0);// };ReRooting<mint> rr(n,op,id);rep(i,n-1){int a,b; cin>>a>>b;a--; b--;rr.add_edge(a,b);}rr.solve();outm;return;}int main(){ios::sync_with_stdio(false);cin.tie(0);int t = 1;//cin>>t;rep(i,t){solve();}return 0;}