# | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
---|---|---|---|---|---|---|---|
566938 | jamezzz | Jail (JOI22_jail) | C++17 | 8 ms | 256 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
using namespace std;
#ifdef DEBUG
#define dbg(...) printf(__VA_ARGS__);
#define getchar_unlocked getchar
#else
#define dbg(...)
#endif
#define sf scanf
#define pf printf
#define fi first
#define se second
#define pb push_back
#define sz(x) (int)x.size()
#define mnto(x,y) x=min(x,(__typeof__(x))y)
#define mxto(x,y) x=max(x,(__typeof__(x))y)
#define INF 1023456789
#define LINF 1023456789123456789
#define all(x) x.begin(), x.end()
#define disc(x) sort(all(x));x.resize(unique(all(x))-x.begin());
typedef long long ll;
typedef long double ld;
typedef pair<int, int> ii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> iii;
typedef tuple<int, int, int, int> iiii;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<pll> vll;
mt19937 rng(time(0));
#define mod 1000000007
inline int add(int a,int b){
int r=a+b;
while(r>=mod)r-=mod;
while(r<0)r+=mod;
return r;
}
inline int mult(int a,int b){
return (int)(((ll)(a*b))%mod);
}
inline int rd(){
int x=0;
char ch=getchar_unlocked();
while(!(ch&16))ch=getchar();//keep reading while current character is whitespace
while(ch&16){//this will break when ‘\n’ or ‘ ‘ is encountered
x=(x<<3)+(x<<1)+(ch&15);
ch=getchar_unlocked();
}
return x;
}
#define maxn 120000
int pfx[maxn];
int main(){
int tc;sf("%d",&tc);
while(tc--){
int n;sf("%d",&n);
for(int i=0;i<n-1;++i){
int a,b;sf("%d%d",&a,&b);
}
int m;sf("%d",&m);
vector<ii> l,r;
for(int i=0;i<m;++i){
int s,t;sf("%d%d",&s,&t);
if(s<t)l.pb({s,t});
else r.pb({t,s});
}
for(int i=1;i<=n;++i)pfx[i]=0;
sort(all(l));sort(all(r));
for(ii pr:r)++pfx[pr.fi];
for(int i=1;i<=n;++i)pfx[i]+=pfx[i-1];
bool die=false;
for(ii pr:l){
if(pfx[pr.se]-pfx[pr.fi-1]!=0)die=true;
}
int far=0;
for(ii pr:l){
if(pr.se<=far)die=true;
far=pr.se;
}
for(ii pr:r)swap(pr.fi,pr.se);
sort(all(r),greater<ii>());
far=n+1;
for(ii pr:r){
if(far<=pr.se)die=true;
far=pr.se;
}
if(die)pf("No\n");
else pf("Yes\n");
}
}
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