2022年11月9日

每日一题-区间覆盖

摘要: 区间覆盖 sort(a.begin(), a.end()); int ans = 0, las = s; for (int i = 0; i < n; ++i) { int j = i; int r = -2e9; while (j < n and a[j].first <= las) { r = 阅读全文

posted @ 2022-11-09 21:53 Whosedream-0019 阅读(30) 评论(0) 推荐(0)

每日一题-区间分组

摘要: 区间分组 sort(a.begin(), a.end()); priority_queue<int, vector<int>, greater<int>> q; for (int i = 0; i < n; ++i) { if (q.empty() or q.top() >= a[i].first) 阅读全文

posted @ 2022-11-09 21:52 Whosedream-0019 阅读(42) 评论(0) 推荐(0)

每日一题-叠罗汉的牛

摘要: 叠罗汉的牛 sort(a.begin(), a.end(), [](const auto &A, const auto &B) { return A.first + A.second < B.first + B.second; }); int sum = 0, ans = -2e9; for (in 阅读全文

posted @ 2022-11-09 21:49 Whosedream-0019 阅读(44) 评论(0) 推荐(0)

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