제출 #203749

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
203749		dimash241	Two Antennas (JOI19_antennas)	C++17	13 / 100	324 ms	46584 KiB

antennas

//#pragma GCC target("avx2")
//#pragma GCC optimize("O3")

//# include <x86intrin.h>
# include <bits/stdc++.h>

# include <ext/pb_ds/assoc_container.hpp>
# include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
using namespace std;
 
template<typename T> using ordered_set = tree <T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

#define _USE_MATH_DEFINES_
#define ll long long
#define ld long double
#define Accepted 0
#define pb push_back
#define mp make_pair
#define sz(x) (int)(x.size())
#define every(x) x.begin(),x.end()
#define F first
#define S second
#define lb lower_bound
#define ub upper_bound
#define For(i,x,y)  for (ll i = x; i <= y; i ++) 
#define FOr(i,x,y)  for (ll i = x; i >= y; i --)
#define SpeedForce ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0)
// ROAD to...                                                                                                                                                                                                                Red

inline void Input_Output () {
	//freopen(".in", "r", stdin);
   //freopen(".out", "w", stdout);
}

const double eps = 0.000001;
const ld pi = acos(-1);
const int maxn = 1e7 + 9;
const int mod = 1e9 + 7;
const ll MOD = 1e18 + 9;
const ll INF = 1e18 + 123;
const int inf = 2e9 + 11;
const int mxn = 1e6 + 9;
const int N = 6e5 + 123;                                          
const int M = 22;
const int pri = 997;
const int Magic = 2101;

const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, -1, 0, 1};
 
int n, m, k;
struct ana {
	int h, a, b;
} a[N];
int dp[2002][2002];


int main () {
   	SpeedForce;
	cin >> n;
	for (int i = 1; i <= n; i ++) {
		cin >> a[i].h >> a[i].a >> a[i].b;
	}
	for (int i = 0; i <= n; i ++)
		for (int j = 0; j <= n; j ++)
			dp[i][j] = -1;
	for (int j = 1; j <= n; ++j) {
		
		for (int k = j + a[j].a; k <= n && k <= j + a[j].b; ++k) if ((k - j) >= a[k].a && a[k].b >= (k - j)) {
			dp[j][k] = abs(a[j].h - a[k].h);
				
		}
	}

	for (int i = n; i >= 1; i --) {
		for (int j = i; j <= n; ++j) {
			dp[i][j] = max(dp[i][j], dp[i][j-1]);
			dp[i-1][j] = max(dp[i-1][j], dp[i][j]);
		}
	}

	cin >> m;
	for (int i = 1; i <= m; i ++) {
		int l, r;
		cin >> l >> r;
		

		cout << dp[l][r] << '\n';
	}

   	return Accepted;
}

// B...a

• Subtask 1: 2 / 2
• Subtask 2: 11 / 11
• Subtask 3: 0 / 22
• Subtask 4: 0 / 65