欢迎来到起遇信息学
起遇信息学正处于上线筹建阶段,以下功能已全部开放免费体验: ✅ 完整题库浏览与代码提交评测(C / C++ / Python / Java 等) ✅ 入门到进阶的系列课程试读、作业与考试 ✅ AI 提示、AI 作业分析等智能助教功能 ✅ 赛事模拟与个人能力报告 ✅ 邮箱注册开放 ⏳ 付费课程订阅与微信/支付宝支付通道 ⏳ 手机号登录,微信扫码登录、微信公众号绑定 使用中如遇任何问题,欢迎通过页面底部 **"联系我们"** 与我们沟通。
CF1795B.Ideal Point
Ideal Point
You are given one-dimensional segments (each segment is denoted by two integers — its endpoints).
Let's define the function as the number of segments covering point (a segment covers the point if , where is the left endpoint and is the right endpoint of the segment).
An integer point is called ideal if it belongs to more segments than any other integer point, i. e. is true for any other integer point .
You are given an integer . Your task is to determine whether it is possible to remove some (possibly zero) segments, so that the given point becomes ideal.
Input
The first line contains one integer () — the number of test cases.
The first line of each test case contains two integers and ().
Then lines follow, -th line of them contains two integers and (; ) — the endpoints of the -th segment.
Output
For each test case, print YES if it is possible to remove some (possibly zero) segments, so that the given point becomes ideal, otherwise print NO.
You may print each letter in any case (YES, yes, Yes will all be recognized as positive answer, NO, no and nO will all be recognized as negative answer).
Note
In the first example, the point is already ideal (it is covered by three segments), so you don't have to delete anything.
In the fourth example, you can delete everything except the segment .
Samples
4
4 3
1 3
7 9
2 5
3 6
2 9
1 4
3 7
1 3
2 4
3 5
1 4
6 7
5 5
YES
NO
NO
YES
在线编程 IDE
建议全屏模式获得最佳体验
| 进入全屏编程 | Alt+E |
| 递交评测 | Ctrl+Enter |
| 注释/取消注释 | Ctrl+/ |
| 缩放字体 | Ctrl+滚轮 |