제출 #536798

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
536798		kym	Railway Trip 2 (JOI22_ho_t4)	C++14	0 / 100	189 ms	59712 KiB

Main

#include <bits/stdc++.h>
using namespace std;
#define int ll 
#define FOR(i,s,e) for(ll i = s; i <= (ll)e; ++i)
#define DEC(i,s,e) for(ll i = s; i >= (ll)e; --i)
#define IAMSPEED ios_base::sync_with_stdio(false); cin.tie(0);
#ifdef LOCAL
#define db(x) cerr << #x << "=" << x << "\n"
#define db2(x, y) cerr << #x << "=" << x << " , " << #y << "=" << y << "\n"
#define db3(a,b,c) cerr<<#a<<"="<<a<<","<<#b<<"="<<b<<","<<#c<<"="<<c<<"\n"
#define dbv(v) cerr << #v << ":"; for (auto ite : v) cerr << ite << ' '; cerr <<"\n"
#define dbvp(v) cerr << #v << ":"; for (auto ite : v) cerr << "{"  << ite.f << ',' << ite.s << "} "; cerr << "\n"
#define dba(a,ss,ee) cerr << #a << ":"; FOR(ite,ss,ee) cerr << a[ite] << ' '; cerr << "\n"
#define reach cerr << "LINE: " << __LINE__ << "\n";
#else
#define db(x)
#define db2(x,y)
#define db3(a,b,c)
#define dbv(v)
#define dbvp(v)
#define dba(a,ss,ee)
#define reach 
#endif
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define ll long long 
#define pb push_back
#define eb emplace_back
#define all(x) (x).begin(), (x).end()
#define f first
#define s second
#define g0(x) get<0>(x)
#define g1(x) get<1>(x)
#define g2(x) get<2>(x)
#define g3(x) get<3>(x)
typedef pair <ll, ll> pi;
typedef tuple<ll,ll,ll> ti3;
typedef tuple<ll,ll,ll,ll> ti4;
ll rand(ll a, ll b) { return a + rng() % (b-a+1); }
const int MOD = 1e9 + 7;
const int inf = (int)1e9 + 500;
const long long oo = (ll)1e18 + 500;
template <typename T> void chmax(T& a, const T b) { a=max(a,b); }
template <typename T> void chmin(T& a, const T b) { a=min(a,b); }
const int MAXN = 100005;
int n,k,m;
pi A[MAXN];
int Q;
struct node {
	int s,e,m,val,lazymx;
	node *l, *r;
	node (int _s, int _e) {
		s=_s,e=_e;
		m=(s+e)/2;
		lazymx=oo;
		if(s!=e){
			l=new node(s,m);
			r=new node(m+1,e);
		}
	}
	void set(int x, int nval) {
		if(s==e){
			val=nval;
			return;
		} else {
			if(x>m)r->set(x,nval);
			else l->set(x,nval);
			val=max(l->val,r->val);
		}
	}
	void value() {
		if(s==e) {
			if(lazymx==oo)return;
			val=max(val,lazymx);
			lazymx=oo;
			return;
		}
		if(lazymx != oo){
			chmax(l->val,lazymx);
			chmax(r->val,lazymx);
			lazymx=oo;
			return;
		}
	}
	void range_set(int x, int y, int nval) {
		value();
		if(s==x&&e==y) {
			lazymx=nval;
			return;
		}
		if(x>m)r->range_set(x,y,nval);
		else if(y<=m)l->range_set(x,y,nval);
		else l->range_set(x,m,nval), r->range_set(m+1,y,nval);
		l->value(); r->value();
		val=max(l->val,r->val);
	}
	int query(int x) {
		value();
		if(s==e) return val;
		if(x>m)return r->query(x);
		else return l->query(x);
	}
} *normal, *back;

int fronttk[MAXN][21];
int backttk[MAXN][21];

int32_t main() 
{
	IAMSPEED
	cin >> n >> k >> m;
	FOR(i,1,m) cin>> A[i].f >> A[i].s;
	normal = new node(1,n);
	back = new node(1,n);
	FOR(i,1,n) {
		normal->set(i,i);
		back->set(i,-i);
	}
	FOR(i,1,m) {
		if(A[i].f < A[i].s) {
			normal->range_set(A[i].f, min(A[i].f + k - 1, A[i].s-1), A[i].s);
		} else {
			back->range_set(max(A[i].f-k+1,A[i].s+1), A[i].f, -A[i].s);
		}
	}
	memset(fronttk,-1,sizeof fronttk);
	memset(backttk,-1,sizeof backttk);
	FOR(i,1,n) {
		fronttk[i][0] = normal->query(i);
		backttk[i][0] = -back->query(i);
	}
	DEC(i,n,1) {
		FOR(j,1,20) {
			if(fronttk[i][j-1]==-1)break;
			fronttk[i][j] = fronttk[fronttk[i][j-1]][j-1];
		}
	}
	FOR(i,1,n) {
		FOR(j,1,20) {
			if(backttk[i][j-1]==-1)break;
			backttk[i][j] = backttk[backttk[i][j-1]][j-1];
			
		}
	}
	int Q; cin >> Q;
	while(Q--) {
		int x, y; cin >> x >> y;
		if(x<y) {
			int ans =0 ;
			DEC(i,20,0) {
				if(fronttk[x][i] != -1 && fronttk[x][i] < y){
					ans += 1ll<<i;
					x=fronttk[x][i];
				}
			}
			++ans;
			if(ans==2097152||fronttk[x][0]<y)ans=-1;
			cout << ans << '\n';
		} else {
			int ans =0 ;
			DEC(i,20,0) {
				if(backttk[x][i] != -1 && backttk[x][i] > y){
					ans += 1ll<<i;
					x=backttk[x][i];
				}
			}
			++ans;
			if(ans==2097152||backttk[x][0]>y)ans=-1;
			cout << ans << '\n';
		}
	}
}

• Subtask 1: 0 / 8
• Subtask 2: 0 / 8
• Subtask 3: 0 / 11
• Subtask 4: 0 / 25
• Subtask 5: 0 / 35
• Subtask 6: 0 / 13