제출 #1290935

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1290935		trvhung	운세 보기 2 (JOI14_fortune_telling2)	C++20	0 / 100	1 ms	332 KiB

fortune_telling2

#include <bits/stdc++.h>
// #include <ext/rope>
// #include <ext/pb_ds/assoc_container.hpp>

// using namespace __gnu_pbds;
// using namespace __gnu_cxx;
using namespace std;

// #define   ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define            ll long long
#define           ull unsigned long long
#define            ld long double
#define            pb push_back
#define  bit(mask, i) ((mask >> i) & 1)
#define            el '\n'
#define             F first
#define             S second

template <class X, class Y> bool maximize(X &x, const Y &y) { return (x < y ? x = y, 1 : 0); }
template <class X, class Y> bool minimize(X &x, const Y &y) { return (x > y ? x = y, 1 : 0); }

const int INF = 1e9;
const ll LINF = 1e18;
const int MOD = 1e9 + 7;
const int MULTI = 0;
const ld eps = 1e-9;
const int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0}; // R D L U
const int ddx[4] = {-1, 1, 1, -1}, ddy[4] = {1, 1, -1, -1}; // UR DR DL UL
const char cx[4] = {'R', 'D', 'L', 'U'};
const ll base = 31;
const int nMOD = 2;
const ll mods[] = {(ll)1e9 + 10777, (ll)1e9 + 19777, (ll)1e9 + 3, (ll)1e9 + 3777};

const int N = 2e5 + 5;
const int LOG = 17;

int n, k, st[LOG + 1][N];
pair<int, int> p[N], t[N];

void preprocess() {
	for (int i = 1; i <= n; ++i) st[0][i] = t[i].S;
	for (int j = 1; j <= LOG; ++j)
		for (int i = 1; i + (1 << j) - 1 <= n; ++i) 
			st[j][i] = max(st[j - 1][i], st[j - 1][i + (1 << (j - 1))]);
}

int queryMax(int l, int r) {
	int k = __lg(r - l + 1);
	return max(st[k][l], st[k][r - (1 << k) + 1]);
}

struct BIT {
	int bit[N];

	void update(int id, int v) {
		for (; id > 0; id -= id & -id)
			bit[id] += v;
	}

	int get(int id) {
		int res = 0;
		for (; id <= n; id += id & -id)
			res += bit[id];
		return res;
	}
} BIT;

void solve() {
	cin >> n >> k;
	for (int i = 1; i <= n; ++i)
		cin >> p[i].F >> p[i].S;
	for (int i = 1; i <= k; ++i) {
		cin >> t[i].F;
		t[i].S = i;
	}

	sort(t + 1, t + 1 + k);
	sort(p + 1, p + 1 + n, [&](pair<int, int> A, pair<int, int> B){
		return max(A.F, A.S) > max(B.F, B.S);
	});

	preprocess();

	int ptr = k;
	ll res = 0;

	for (int i = 1; i <= n; ++i) 
		if (p[i].F == p[i].S)
			res += p[i].F;
		else {
			while (ptr >= 1 && t[ptr].F >= max(p[i].F, p[i].S)) 
				BIT.update(t[ptr--].S, 1);

			int mn = min(p[i].F, p[i].S), mx = max(p[i].F, p[i].S);
			int l = lower_bound(t + 1, t + 1 + n, make_pair(mn, 0)) - t;
			int r = upper_bound(t + 1, t + 1 + n, make_pair(mx - 1, n)) - t - 1;

			int lst = 0;
			if (l <= r && r <= n) lst = queryMax(l, r);

			int cnt = BIT.get(lst + 1);
			if (lst == 0) {
				if (cnt & 1)
					res += p[i].S;
				else
					res += p[i].F;
			} else {
				if (cnt & 1)
					res += mn;
				else
					res += mx;
			}
		}

	cout << res;
}

signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

    if (!MULTI) solve();
    else {
        int test; cin >> test;
        while (test--) solve();
    }
    
    return 0;
}

• Subtask 1: 0 / 4
• Subtask 2: 0 / 31
• Subtask 3: 0 / 65