【Luogu P6902】[ICPC2014 WF] Surveillance(倍增,贪心)

该博客介绍了如何解决一个国际保护和控制组织(ICPC)的监控问题,即如何用最少的相机覆盖其总部建筑的所有墙壁。问题转换为在环形序列上寻找最少的线段来覆盖至少一半的序列。博客提出了使用贪心策略结合倍增算法优化,将时间复杂度降低到O(nlogn)的方法,并给出了详细的思路和代码实现。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

题目

题目描述

The International Corporation for Protection and Control (ICPC) develops efficient technology for, well, protection and control. Naturally, they are keen to have their own headquarters protected and controlled. Viewed from above, the headquarters building has the shape of a convex polygon. There are several suitable places around it where cameras can be installed to monitor the building. Each camera covers a certain range of the polygon sides (building walls), depending on its position. ICPC wants to minimize the number of cameras needed to cover the whole building.

输入格式

The input consists of a single test case. Its first line contains two integers n n n and k k k ( 3 ≤ n ≤ 1 0 6 3 \le n \le 10^6 3n106 and 1 ≤ k ≤ 1 0 6 1 \le k \le 10^6 1k106), where n n n is the number of walls and k k k is the number of possible places for installing cameras. Each of the remaining k k k lines contains two integers a i a_ i ai and b i b_ i bi ( 1 ≤ a i , b i ≤ n 1 \le a_ i, b_ i \le n 1ai,bin). These integers specify which walls a camera at the i t h i^{th} ith place would cover. If

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值