제출 #1124921

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1124921		pan	Advertisement 2 (JOI23_ho_t2)	C++20	59 / 100	223 ms	86596 KiB

Main

#include <bits/stdc++.h>
//#include "includeall.h"
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define endl '\n'
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define input2(x, y) scanf("%lld%lld", &x, &y);
#define input3(x, y, z) scanf("%lld%lld%lld", &x, &y, &z);
#define input4(x, y, z, a) scanf("%lld%lld%lld%lld", &x, &y, &z, &a);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define all(x) x.begin(), x.end()
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
#define FOR(i, x, n) for (ll i =x; i<=n; ++i) 
#define RFOR(i, x, n) for (ll i =x; i>=n; --i) 
using namespace std;
//using namespace __gnu_pbds;
//#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
//#define ordered_multiset tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
typedef long long ll;
//typedef __int128 ull;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<pi, ll> pii;
typedef pair<pi, pi> piii;

ll const INF = 1e13;

struct node {
    int s, e, m; 
    pi minn, maxx;
    node * l, * r; 
    node(int S, int E) { 
        s = S, e = E, m = (s + e) / 2;
        minn = maxx = mp(0, s);
        if (s != e) { 
            l = new node(s, m); 
            r = new node(m + 1, e); 
        }
    }
    void update(int X, ll m1,ll m2) { 
        if (s == e) minn = mp(m1, s), maxx = mp(m2, s);
        else { 
            if (X <= m) l -> update(X, m1, m2); 
            else r -> update(X, m1, m2); 
            minn = min(l->minn, r->minn);
            maxx = max(l->maxx, r-> maxx);
        }
    }
    pi querymin(int S, int E) {
		if (S > E) return mp(INF, -1);
        if (s == S && e == E) return minn;
        else if (E <= m) return l -> querymin(S, E);
        else if (S >= m + 1) return r -> querymin(S, E);
        else return min(l -> querymin(S, m), r -> querymin(m + 1, E));
    }
    pi querymax(int S, int E) {
		if (S > E) return mp(-INF, -1);
        if (s == S && e == E) return maxx;
        else if (E <= m) return l -> querymax(S, E);
        else if (S >= m + 1) return r -> querymax(S, E);
        else return max(l -> querymax(S, m), r -> querymax(m + 1, E));
    }
}* root;

ll x[200005], e[200005];
bool die[200005];

int main()
{
	ll n;
	cin >> n;
	root = new node(1, n);
	vector<pi> v, v2(n);
	for (ll i=0; i<n; ++i) 
	{
		input2(v2[i].f, v2[i].s);
	}
	sort(all(v2));
	for (ll i=1; i<=n; ++i) 
	{
		x[i] = v2[i-1].f, e[i] = v2[i-1].s;
		die[i] = 0;
		root->update(i, x[i] + e[i], x[i] - e[i]);
		v.pb(mp(e[i], i));
		//show2(x[i], e[i]);
	}
	sort(all(v), greater<pi>());
	ll ans = 0;
	for (ll i=0; i<n; ++i)
	{
		ll idx = v[i].s;
		if (die[idx]) continue;
		//show(idx);
		die[idx] = 1;
		ans++;
		root->update(idx, INF, -INF);
		while (1)
		{
			pi cur = root->querymax(1, idx-1);
			if (cur.f < x[idx] - e[idx]) break;
			die[cur.s] = 1;
			root->update(cur.s, INF, -INF);
		}
		while (1)
		{
			pi cur = root->querymin(idx+1, n);
			if (cur.f > x[idx] + e[idx]) break;
			die[cur.s] = 1;
			root->update(cur.s, INF, -INF);
		}
		
	}
	cout << ans << endl;
	
	
	return 0;
}

컴파일 시 표준 에러 (stderr) 메시지

Main.cpp: In function 'int main()':
Main.cpp:13:27: warning: ignoring return value of 'int scanf(const char*, ...)' declared with attribute 'warn_unused_result' [-Wunused-result]
   13 | #define input2(x, y) scanf("%lld%lld", &x, &y);
      |                      ~~~~~^~~~~~~~~~~~~~~~~~~~
Main.cpp:87:17: note: in expansion of macro 'input2'
   87 |                 input2(v2[i].f, v2[i].s);
      |                 ^~~~~~

• Subtask 1: 0 / 10
• Subtask 2: 23 / 23
• Subtask 3: 36 / 36
• Subtask 4: 0 / 31