codeforces 1288DMinimax Problem

題意:n(3e5)個數組,每個數組m個數(m<8),兩個數組合並規則爲對m個數每個數取較大的數,求合併之後數組最小值的最大的兩個數組

二分合並之後的最小值,然後枚舉(1<<m)^2次,判斷是否滿足條件

#include <iostream>
constexpr int N = 300'000, M = 8;
int n, m, x, y, tot;
int a[N][M], id[1 << M];
bool solve(int v) {
	std::fill(id, id + tot, -1);
	for (int i = 0; i < n; ++i) {
		int s = 0;
		for (int j = 0; j < m; ++j)
			if (a[i][j] >= v)
				s |= 1 << j;
		id[s] = i;
	}
	for (int i = tot - 1; i >= 0; --i)
		for (int j = 0; j < m; ++j)
			if (id[i | 1 << j] != -1)
				id[i] = id[i | 1 << j];
	for (int i = 0; i < tot; ++i) {
		if (id[i] != -1 && id[(tot - 1) ^ i] != -1) {
			x = id[i];
			y = id[(tot - 1) ^ i];
			return true;
		}
	}
	return false;
}
int main() {
	std::ios::sync_with_stdio(false);
	std::cin.tie(nullptr);
	std::cin >> n >> m;
	tot = 1 << m;
	for (int i = 0; i < n; ++i)
		for (int j = 0; j < m; ++j)
			std::cin >> a[i][j];
	int l = 0, r = 1'000'000'000;
	while (l < r) {
		int m = (l + r + 1) / 2;
		if (solve(m)) {
			l = m;
		}
		else {
			r = m - 1;
		}
	}
	solve(l);
	std::cout << x + 1 << " " << y + 1 << "\n";
	return 0;
}