제출 #1308253

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1308253		Nonoze	Pairs (IOI07_pairs)	C++17	100 / 100	146 ms	18280 KiB

pairs

/*
*	Author: Nonoze
*	Created: Sunday 04/01/2026
*/
#include <bits/stdc++.h>
using namespace std;

#ifndef DEBUG
	#define dbg(...)
#endif

// #define cout cerr << "OUT: "
#define endl '\n'
#define endlfl '\n' << flush
#define quit(x) return (void)(cout << x << endl)

template<typename T> void read(T& x) { cin >> x; }
template<typename T1, typename T2> void read(pair<T1, T2>& p) { read(p.first), read(p.second); }
template<typename T> void read(vector<T>& v) { for (auto& x : v) read(x); }
template<typename T1, typename T2> void read(T1& x, T2& y) { read(x), read(y); }
template<typename T1, typename T2, typename T3> void read(T1& x, T2& y, T3& z) { read(x), read(y), read(z); }
template<typename T1, typename T2, typename T3, typename T4> void read(T1& x, T2& y, T3& z, T4& zz) { read(x), read(y), read(z), read(zz); }
template<typename T> void print(vector<T>& v) { for (auto& x : v) cout << x << ' '; cout << endl; }

#define sz(x) (int)(x.size())
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define make_unique(v) sort(all(v)), v.erase(unique(all(v)), (v).end())
#define pb push_back
#define mp(a, b) make_pair(a, b)
#define fi first
#define se second
#define cmin(a, b) a = min(a, b)
#define cmax(a, b) a = max(a, b)
#define YES cout << "YES" << endl
#define NO cout << "NO" << endl
#define QYES quit("YES")
#define QNO quit("NO")

#define int long long
#define double long double
const int inf = numeric_limits<int>::max() / 4;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int MOD = 1e9+7, LOG=20;



void solve();

signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	int tt=1;
	// cin >> tt;
	while(tt--) solve();
	return 0;
}




template<class T>
struct segtree {
	vector<T> st;
	int n;
	T idElement;

	segtree() {}

	segtree(int nn, T idEl, vector<T> creation={}) {
		n=nn, idElement=idEl;
		st.resize(n*2, idElement);
		if (!creation.empty()) {
			for (int i=0; i<n; i++) st[i+n]=creation[i];
			build();
		}
	}

	void build() {
		for (int i=n-1; i>0; i--) st[i] = combine(st[i<<1], st[i<<1|1]);
	}
	
	void update(int p, T val) {
		for (st[p += n] += val; p > 1; p >>= 1) st[p>>1] = combine(st[p], st[p^1]);
	}

	T query(int l, int r) {
		T resl = idElement, resr = idElement;
		r++;
		for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
			if (l&1) resl = combine(resl, st[l++]);
			if (r&1) resr = combine(st[--r], resr);
		}
		return combine(resl, resr);
	}

	T combine(T l, T r) {
		T ans=l+r;
		return ans;
	}
};



struct grid {
	int x, y, z;
	int a, b;
};

int b, n, k, m;


int sub1() {
	vector<int> a(n);
	for (auto &u: a) cin >> u;
	sort(all(a));
	int pos=1, ans=0;
	for (int i=0; i<n; i++) {
		pos=max(pos, i+1);
		while (pos<n && a[pos]-a[i]<=k) pos++;
		pos--;
		ans+=pos-i;
	}
	return ans;
}


int sub2() {
	vector<grid> a(n);
	for (auto &u: a) cin >> u.x >> u.y, --u.x, --u.y, u.a=u.x+u.y, u.b=u.x-u.y+m;

	vector<vector<grid>> buckets(2*m+1); for (auto &u: a) buckets[u.a].pb(u);
	segtree<int> st(2*m+1, 0);
	int ans=0;
	for (int i=0; i<=2*m; i++) {
		if (i>k) for (auto &u: buckets[i-k-1]) st.update(u.b, -1);
		for (auto &u: buckets[i]) {
			assert(u.b>=0 && u.b<=2*m);
			int low=max(0LL, u.b - k);
			int high=min(2*m, u.b + k);
			int res=st.query(low, high);
			ans+=res;
			st.update(u.b, 1);
		}
	}
	return ans;
}


int sub3() {
	if (k>=3*m) return n*(n-1)/2;
	vector<grid> a(n);
	for (auto &u: a) {
		cin >> u.x >> u.y >> u.z; u.x--, u.y--, u.z--;
		u.a=u.y+u.z;
		u.b=u.y-u.z + m;
	}

	vector<vector<vector<int>>> prefix(m, vector<vector<int>>(2*m+1, vector<int>(2*m+1, 0)));
	for (auto &u: a) prefix[u.x][u.a][u.b]++;
	for (int x=0; x<m; x++) {
		for (int i=0; i<=2*m; i++) {
			for (int j=0; j<=2*m; j++) {
				if (j>0) prefix[x][i][j] += prefix[x][i][j-1];
				if (i>0) prefix[x][i][j] += prefix[x][i-1][j];
				if (i>0 && j>0) prefix[x][i][j] -= prefix[x][i-1][j-1];
			}
		}
	}

	int ans=0;

	for (auto &u: a) {
		int cnt=0;
		for (int x= max(0LL, u.x - k); x<= min(m-1, u.x + k); x++) {
			int lowa = max(0LL, u.a - (k - abs(u.x - x))), higha = min(2*m, u.a + (k - abs(u.x - x)));
			int lowb = max(0LL, u.b - (k - abs(u.x - x))), highb = min(2*m, u.b + (k - abs(u.x - x)));
			int res=prefix[x][higha][highb];
			if (lowa>0) res -= prefix[x][lowa-1][highb];
			if (lowb>0) res -= prefix[x][higha][lowb-1];
			if (lowa>0 && lowb>0) res += prefix[x][lowa-1][lowb-1];
			cnt+=res;
		}
		ans+=cnt-1;
	}
	return ans/2;
}

void solve() {
	cin >> b >> n >> k >> m;
	cout << (b==1 ? sub1() : (b==2 ? sub2() : sub3())) << endl;
	return;
}

• Subtask 1: 6 / 6
• Subtask 2: 6 / 6
• Subtask 3: 6 / 6
• Subtask 4: 6 / 6
• Subtask 5: 6 / 6
• Subtask 6: 7 / 7
• Subtask 7: 7 / 7
• Subtask 8: 8 / 8
• Subtask 9: 8 / 8
• Subtask 10: 10 / 10
• Subtask 11: 10 / 10
• Subtask 12: 10 / 10
• Subtask 13: 10 / 10