結果

問題 No.363 門松サイクル
ユーザー koyumeishi
提出日時 2016-03-16 01:49:45
言語 C++11(廃止可能性あり)
(gcc 13.3.0)
結果
AC  
実行時間 623 ms / 4,000 ms
コード長 3,570 bytes
コンパイル時間 1,298 ms
コンパイル使用メモリ 104,304 KB
実行使用メモリ 36,096 KB
最終ジャッジ日時 2024-10-01 20:09:01
合計ジャッジ時間 12,804 ms
ジャッジサーバーID
(参考情報)
judge5 / 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 * 2
other AC * 27
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 <iostream>
#include <vector>
#include <cstdio>
#include <sstream>
#include <map>
#include <string>
#include <algorithm>
#include <queue>
#include <cmath>
#include <functional>
#include <set>
#include <ctime>
#include <random>
#include <chrono>
#include <cassert>
using namespace std;
//input from stdin
template<class T> istream& operator >> (istream& is, vector<T>& vec){for(T& val: vec) is >> val; return is;}
template<class T> istream& operator , (istream& is, T& val){ return is >> val;}
template<class T> ostream& operator << (ostream& os, const vector<T>& vec){for(int i=0; i<vec.size(); i++) os << vec[i] << (i==vec.size()-1?"":" ");
    return os;}
template<class T> ostream& operator , (ostream& os, const T& val){ return os << " " << val;}
template<class T> ostream& operator >> (ostream& os, const T& val){ return os << " " << val;}
bool is_kadomatsu(vector<int> v){
if(v.size()!=3) return false;
if(v[0]==v[1] || v[1]==v[2] || v[2]==v[0]) return false;
return (v[0]<v[1] && v[1]>v[2]) || (v[0]>v[1] && v[1]<v[2]);
}
int main_(){
int n;
cin >> n;
vector<int> a(n);
cin >> a;
vector<vector<int>> G(n);
for(int i=0; i<n-1; i++){
int u,v; cin >> u,v;
//scanf("%d%d", &u,&v);
u--,v--;
G[u].push_back(v);
G[v].push_back(u);
}
vector<int> depth(n, -1);
vector<int> kadomatsu(n, 0);
vector<vector<int>> par(n, vector<int>(20));
function<void(int,int,int)> dfs = [&](int pos, int last, int last2){
kadomatsu[pos] = is_kadomatsu({a[pos],a[last],a[last2]});
depth[pos] = depth[last]+1;
par[pos][0] = last;
for(int next : G[pos]){
if(next == last) continue;
dfs(next, pos, last);
}
};
dfs(0,0,0);
for(int k=1; k<20; k++){
for(int i=0; i<n; i++){
par[i][k] = par[ par[i][k-1] ][k-1];
kadomatsu[i] |= ( kadomatsu[i] & kadomatsu[ par[i][k-1] ] & (1<<(k-1)) ) << 1;
}
}
auto kth_parent = [&](int pos, int k){
if(k<0) return -1;
for(int i=0; k!=0; k>>=1, i++) if(k&1) pos = par[pos][i];
return pos;
};
auto lca = [&](int x, int y){
if(depth[x] < depth[y]) swap(x,y);
x = kth_parent(x, depth[x] - depth[y]);
if(x==y) return x;
for(int k=19; k>=0; k--){
if(par[x][k] != par[y][k]){
x = par[x][k];
y = par[y][k];
}
}
return par[x][0];
};
vector<int> log_k(n+1, 0);
for(int i=2, tmp=0; i<log_k.size(); i++){
log_k[i] = tmp;
if((i&-i) == i) tmp++;
}
auto is_kadomatsu_line = [&](int x, int y){
if(depth[x] < depth[y]) swap(x,y);
//assert(lca(x,y) == y);
int d = depth[x] - depth[y] - 2;
if(d<0) return true;
int k = log_k[d+1];
y = kth_parent(x, d-(1<<k)+1);
int ok = kadomatsu[x] & kadomatsu[y] & (1<<k);
return ok>0;
};
int q;
cin >> q;
while(q--){
int u,v;
//scanf("%d%d", &u,&v);
cin >> u,v;
u--,v--;
assert(u!=v);
int p = lca(u,v);
bool ok = true;
if(p==u || p==v){ //straight line
if(p==u) swap(u,v); // v==p
int pu = kth_parent(u, depth[u]-depth[p]-1);
int qu = par[u][0];
ok = is_kadomatsu_line(u,p) && is_kadomatsu({a[qu], a[u], a[p]}) && is_kadomatsu({a[u], a[v], a[pu]});
}else{
int pu = kth_parent(u, depth[u]-depth[p]-1);
int pv = kth_parent(v, depth[v]-depth[p]-1);
int qu = par[u][0];
int qv = par[v][0];
ok = (
is_kadomatsu_line(u,p) &&
is_kadomatsu_line(v,p) &&
is_kadomatsu({a[pu], a[p], a[pv]}) &&
is_kadomatsu({a[qu], a[u], a[v] }) &&
is_kadomatsu({a[u], a[v], a[qv]})
);
}
if(ok) cout << "YES" << "\n";
else cout << "NO" << "\n";
}
return 0;
}
int main(){
//auto s = clock();
main_();
//cerr << clock()-s << " [ms] " << endl;
return 0;
}
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