結果
問題 | No.363 門松サイクル |
ユーザー |
|
提出日時 | 2022-09-10 13:22:41 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
WA
|
実行時間 | - |
コード長 | 5,624 bytes |
コンパイル時間 | 2,958 ms |
コンパイル使用メモリ | 215,488 KB |
最終ジャッジ日時 | 2025-02-07 03:52:21 |
ジャッジサーバーID (参考情報) |
judge5 / judge4 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 2 |
other | AC * 5 WA * 22 |
ソースコード
#include<bits/stdc++.h>using namespace std;#define rep(i,n) for(ll i=0;i<n;i++)#define repl(i,l,r) for(ll i=(l);i<(r);i++)#define per(i,n) for(ll i=(n)-1;i>=0;i--)#define perl(i,r,l) for(ll i=r-1;i>=l;i--)#define fi first#define se second#define pb push_back#define ins insert#define pqueue(x) priority_queue<x,vector<x>,greater<x>>#define all(x) (x).begin(),(x).end()#define CST(x) cout<<fixed<<setprecision(x)#define vtpl(x,y,z) vector<tuple<x,y,z>>#define rev(x) reverse(x);using ll=long long;using vl=vector<ll>;using vvl=vector<vector<ll>>;using pl=pair<ll,ll>;using vpl=vector<pl>;using vvpl=vector<vpl>;const ll MOD=1000000007;const ll MOD9=998244353;const int inf=1e9+10;const ll INF=4e18;const ll dy[9]={0,1,0,-1,1,1,-1,-1,0};const ll dx[9]={1,0,-1,0,1,-1,1,-1,0};template<class T> inline bool chmin(T& a, T b) {if (a > b) {a = b;return true;}return false;}template<class T> inline bool chmax(T& a, T b) {if (a < b) {a = b;return true;}return false;}bool isKad(ll a,ll b,ll c){if(a==b||a==c||b==c)return false;if(a<b&&b<c)return false;if(a>b&&b>c)return false;return true;}struct Edge {long long to;};using Graph = vector<vector<Edge>>;struct LCA {vector<vector<int>> parent; // parent[k][u]:= u の 2^k 先の親vector<int> dist; // root からの距離LCA(const Graph &G, int root = 0) { init(G, root); }void init(const Graph &G, int root = 0) {int V = G.size();int K = 1;while ((1 << K) < V) K++;parent.assign(K, vector<int>(V, -1));dist.assign(V, -1);dfs(G, root, -1, 0);for (int k = 0; k + 1 < K; k++) {for (int v = 0; v < V; v++) {if (parent[k][v] >= 0) {parent[k + 1][v] = parent[k][parent[k][v]];}}}}// 根からの距離と1つ先の頂点を求めるvoid dfs(const Graph &G, int v, int p, int d) {parent[0][v] = p;dist[v] = d;for (auto e : G[v]) {if (e.to != p) {dfs(G, e.to, v, d + 1);}}}int query(int u, int v) {if (dist[u] < dist[v]) swap(u, v); // u の方が深いとするint K = parent.size();// LCA までの距離を同じにするfor (int k = 0; k < K; k++) {if ((dist[u] - dist[v])&(1<<k) ){u = parent[k][u];}}// 二分探索で LCA を求めるif (u == v) return u;for (int k = K - 1; k >= 0; k--) {if (parent[k][u] != parent[k][v]) {u = parent[k][u];v = parent[k][v];}}return parent[0][u];}int length(int u, int v) { return dist[u] + dist[v] - 2 * dist[query(u, v)]; }bool is_in(int u, int v, int a) { return length(u, a) + length(a, v) == length(u, v); }int ancestor(int v,int k){//vからk個根側の頂点番号,存在しないなら-1int K = parent.size();for(int i=K-1;i>=0;i--){if(k>>i&1)v=parent[i][v];if(v==-1)return -1;}return v;}};int main(){ll n;cin >> n;vl a(n);rep(i,n)cin >> a[i];Graph g(n);rep(i,n-1){ll a,b;cin >> a >> b;a--;b--;g[a].push_back({b});g[b].push_back({a});}LCA lc(g);vl dp(n);auto dfs=[&](auto &self,ll v,ll par,ll ppar)->void{if(par!=-1){dp[v]=(isKad(a[v],a[par],a[ppar])? 1:0);dp[v]+=dp[par];}//ll opt=0;for(auto f:g[v]){if(f.to==par)continue;self(self,f.to,v,par);//opt++;}//if(opt==0)dp[v]=dp[par];};dfs(dfs,0,-1,-1);//rep(i,n)cout << dp[i] << endl;ll q;cin >> q;while(q--){ll u,v;cin >> u >> v;u--;v--;ll l=lc.query(u,v);if(u==l||v==l){//cout << "ERROR" << endl;continue;if(lc.length(u,v)==1){cout <<"NO" << endl;continue;}if(u!=l)swap(u,v);ll s=lc.ancestor(v,lc.length(u,v)-1);ll t=lc.ancestor(v,1);if(dp[v]-dp[s]!=lc.length(u,v)-1){cout << "NO" << endl;continue;}if(!isKad(a[s],a[u],a[v])){cout << "NO" << endl;continue;}if(!isKad(a[u],a[v],a[t])){cout << "NO" << endl;continue;}cout << "YES" << endl;}else{ll s=lc.ancestor(u,lc.length(u,l)-1);//cout << lc.length(u,l) << endl;ll t=lc.ancestor(v,lc.length(v,l)-1);//cout << lc.level_ancestor(u,0) << endl;//cout << s << " " << u << endl;ll f=dp[u]-dp[s];if(f!=lc.length(l,u)-1){cout << f << endl;cout << "NO" << endl;continue;}f=dp[v]-dp[t];if(f!=lc.length(l,v)-1){cout << "NO" << endl;continue;}if(!isKad(a[s],a[l],a[t])){cout << "NO" << endl;continue;}s=lc.ancestor(u,1);t=lc.ancestor(v,1);if(!isKad(a[s],a[u],a[v])){cout << "NO" << endl;continue;}if(!isKad(a[u],a[v],a[t])){cout << "NO" << endl;continue;}cout << "YES" << endl;}}}