제출 #681735

#제출 시각아이디문제언어결과실행 시간메모리
681735NK_Fortune Telling 2 (JOI14_fortune_telling2)C++17
100 / 100
262 ms63764 KiB
#include <bits/stdc++.h>

using namespace std;

#define nl '\n'

using ll = long long;

struct BIT {
	vector<int> data; int N;
	void init(int n) { N = n; data = vector<int>(N, 0); }
	void add(int p, int x) { for(++p;p<=N;p+=p&-p) data[p-1] += x; }
	int sum(int l, int r) { return sum(r+1) - sum(l); }
	int sum(int r) { int s = 0; for(;r;r-=r&-r) s += data[r-1]; return s; }
};

struct ST {
	int level(int x) { return 31 - __builtin_clz(x); }
	int comb(int a, int b) { return max(a, b); }
	vector<int> v; vector<vector<int>> jmp;
	void init(const vector<int> &_v) {
		v = _v; jmp = {v};
		for(int j = 1; (1<<j) <= int(size(v)); j++) {
			jmp.push_back(vector<int>(int(size(v)) - (1<<j) + 1));
			for(int i = 0; i < int(size(jmp[j])); i++) {
				jmp[j][i] = comb(jmp[j-1][i], jmp[j-1][i + (1<<(j-1))]);
			}	
		}
	}

	int query(int l, int r) {
		if (l > r) return -1;
		int d = level(r - l + 1);
		return comb(jmp[d][l], jmp[d][r - (1<<d) + 1]);
	}
};

int main() {
	cin.tie(0)->sync_with_stdio(0);
	
	int N, Q; cin >> N >> Q;

	vector<int> A(N), B(N), T(Q);
	vector<int> X;
	for(int i = 0; i < N; i++) {
		cin >> A[i] >> B[i];
		X.push_back(A[i]); X.push_back(B[i]);
	}
	for(int i = 0; i < Q; i++) {
		cin >> T[i];
		X.push_back(T[i]);
	}

	sort(begin(X), end(X));
	X.erase(unique(begin(X), end(X)), end(X));

	vector<int> a = A, b = B;

	int M = size(X);

	auto get = [&](int x) {
		return lower_bound(begin(X), end(X), x) - begin(X);
	};

	for(auto &x : a) x = get(x);
	for(auto &x : b) x = get(x);
	for(auto &x : T) x = get(x);


	vector<int> ti(M, -1); for(int i = 0; i < Q; i++) ti[T[i]] = max(ti[T[i]], i);

	ST st; st.init(ti);

	vector<vector<int>> qry(Q);

	vector<int> S(N, 0);
	for(int i = 0; i < N; i++) {
		int L = min(a[i], b[i]), R = max(a[i], b[i]) - 1;
		int time = st.query(L, R);
		if (time != -1) S[i] = (a[i] >= b[i] ? 0 : 1);
		else time = 0;
		qry[time].push_back(i);
	}

	ll ans = 0;

	BIT F; F.init(M);
	for(int i = Q - 1; i >= 0; i--) {
		F.add(T[i], 1);

		for(auto x : qry[i]) {
			int mx = max(a[x], b[x]);
			int amt = F.sum(mx, M-1);

			S[x] ^= (amt % 2);
			// cout << x << " " << S[x] << nl;

			if (S[x]) ans += B[x];
			else ans += A[x];
			// cout << ans << nl;
		}	
	}	


	cout << ans << nl;

    return 0;
}
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...