时间限制: 1 Sec 内存限制: 128 MB

5G is the proposed next telecommunications standards beyond the current 4G standards. 5G planning aims at higher capacity than current 4G, allowing a higher density of mobile broadband users, and supporting device-to- device, reliable, and massive wireless communications. A telecommunication company would like to install more base stations to provide better communication for customers. Due to the installation cost and available locations,the company can only install S (2 ≤ S ≤ L) base stations at L (2 ≤ L ≤ 100, 000) candidate locations. Since the base stations work in the same frequency band, they will interfere and cause severe performance degradation. To provide high quality communication experience to customers, the company would like to maximize the distance between the base stations so as to reduce the wireless interference among the base stations. Suppose the L candidate locations are in a straight line at locations P1, P2,…, PL (0 ≤ Pi ≤ 1, 000, 000) and the company wants to install S base stations at the L candidate locations. What is the largest minimum distance among the S base stations?

The input data includes multiple test sets.
Each set starts with a line which specifies L (i.e., the number of candidate locations) and S (i.e., the number of base stations). The next line contains L space-separated integers which represent P1, P2,…, PL.
The input data ends “0 0”.
For each set, you need to output a single line which should be the largest minimum distance among the base stations.

sample input
5 3
2 3 9 6 11
4 3
1 4 9 10
0 0

sample output
4

给你L个点,要你选S个点,要求这些点之间的最小距离最大,输出这个最大值
二分答案

#define IN_PC() freopen("C:\\Users\\hz\\Desktop\\in.txt","r",stdin)
#define OUT_PC() freopen("C:\\Users\\hz\\Desktop\\out.txt","w",stdout)
#include <bits/stdc++.h> using namespace std; const int maxn = 100005;
int L,S,P[maxn]; bool judge(int _val) {
int ret = 1;
int las = P[0];
for(int i = 1; i<L; i++) {
if(P[i]-las>=_val) {
ret++;
las = P[i];
}
}
return ret>=S;
} int main() {
// IN_PC();
// OUT_PC();
while(scanf("%d%d",&L,&S)&&!(L==0&&S==0)) {
for(int i=0; i<L; i++) {
scanf("%d",P+i);
}
sort(P,P+L);
int l = 0,r = 1000005;
while(l<r) {
int mid = ((l+r)%2)?((l+r)/2+1):((l+r)/2);
if(judge(mid))
l = mid;
else
r = mid-1;
}
printf("%d\n",(l+r)/2);
}
return 0;
}

【二分】Base Station Sites @ICPC2017HongKong/upcexam5559的更多相关文章

  1. hdu3879 Base Station 最大权闭合子图 边权有正有负

    /** 题目:hdu3879 Base Station 最大权闭合子图 边权有正有负 链接:http://acm.hdu.edu.cn/showproblem.php?pid=3879 题意:给出n个 ...

  2. HDU 3879 Base Station

    Base Station Time Limit: 2000ms Memory Limit: 32768KB This problem will be judged on HDU. Original I ...

  3. 2017 ACM-ICPC 亚洲区(南宁赛区)网络赛 GSM Base Station Identification (点在多边形内模板)

    In the Personal Communication Service systems such as GSM (Global System for Mobile Communications), ...

  4. hdu 3879 Base Station 最大权闭合图

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3879 A famous mobile communication company is plannin ...

  5. HDU 3879 Base Station(最大权闭合子图)

    经典例题,好像说可以转化成maxflow(n,n+m),暂时只可以勉强理解maxflow(n+m,n+m)的做法. 题意:输入n个点,m条边的无向图.点权为负,边权为正,点权为代价,边权为获益,输出最 ...

  6. HDU 3897 Base Station (网络流,最大闭合子图)

    题意:给定n个带权点m条无向带权边,选一个子图,则这个子图的权值为 边权和-点权和,求一个最大的权值. 析:把每条边都看成是一个新点,然后建图,就是一个裸的最大闭合子图. 代码如下: #pragma ...

  7. HDU 3879 Base Station(最大权闭合子图)

    将第i个用户和他需要的基站连边,转化成求二分图的最大权闭合子图. 答案=正权点之和-最小割. # include <cstdio> # include <cstring> # ...

  8. ACdream 1127 Base Station (离线查询+树状数组)

    题目链接: http://acdream.info/problem?pid=1127 题目: 移动通信系统中,通信网的建立主要通过基站来完成. 基站可以分为主基站和子基站.子基站和各个移动用户进行连接 ...

  9. HDU 3879 && BZOJ 1497:Base Station && 最大获利 (最大权闭合图)

    http://acm.hdu.edu.cn/showproblem.php?pid=3879 http://www.lydsy.com/JudgeOnline/problem.php?id=1497 ...

随机推荐

  1. 跨域资源共享 CORS 详解(转自阮一峰博客)

    CORS是一个W3C标准,全称是"跨域资源共享"(Cross-origin resource sharing). 它允许浏览器向跨源服务器,发出XMLHttpRequest请求,从 ...

  2. (二)Makefile——自动编译、清理、安装软件

    每次都要敲击冗长的编译命令,每次都要清理之前的编译中间结果和最终结果,在安装软件时复制剪切大量的动态库,在卸载软件时删除很多的动态库和配置文件.好吧,我被逼向了makefile. helloworld ...

  3. 【转】通过 INotifyPropertyChanged 实现观察者模式

    通过 INotifyPropertyChanged 实现观察者模式 原博客地址 http://frankdzu.blog.sohu.com/117654536.html 普通观察者模式存在的问题 我们 ...

  4. 如何查找物理cpu,cpu核心和逻辑cpu的数量

    环境 Red Hat Enterprise Linux 4 Red Hat Enterprise Linux 5 Red Hat Enterprise Linux 6 Red Hat Enterpri ...

  5. lvs-dr

    第5节  dr模型 在rs上配置 :rip  和vip   vip定义在lo别名上 Director 上配置:vip  和dip 都只需要一块网卡  网卡都桥接 Vip: 192.168.0.105 ...

  6. 2018牛客网暑假ACM多校训练赛(第四场)A Ternary String 数论

    原文链接https://www.cnblogs.com/zhouzhendong/p/NowCoder-2018-Summer-Round4-A.html 题目传送门 - https://www.no ...

  7. BZOJ3926 [Zjoi2015]诸神眷顾的幻想乡 字符串 SAM

    原文链接https://www.cnblogs.com/zhouzhendong/p/BZOJ3926.html 题目传送门 - BZOJ3926 题意 给定一个有 $n$ 个节点,最多只有 $20$ ...

  8. 安卓编程资源文件string中对占位符的使用详解

    这里将为你详细介绍占位符的使用,将其学以致用,可以达到简化布局文件,减少字符串资源量. 1.在资源文件中的使用. 打开资源文件中的strings.xml文件,进行编辑.如下图所示: 图  1.0 2. ...

  9. DDoS攻击与防御(2)

    2.攻击系统资源终端设备在与服务器进行通信时,经常需要创建会话连接,在此过程中通常会使用TCP和SSL等协议.会话连接一旦被占满,新进入的会话请求就必须等待前面的会话完成.消耗系统资源的DDoS攻击的 ...

  10. apache环境配置 | httpd Could not reliably determine the server's fully qualified domain name

    apache环境配置 | httpd Could not reliably determine the server's fully qualified domain name    转 https: ...