题目链接:

Fire-Control System

Time Limit: 12000/5000 MS (Java/Others)   

 Memory Limit: 32768/32768 K (Java/Others)

Problem Description
A new mighty weapon has just been developed, which is so powerful that it can attack a sector of indefinite size, as long as the center of the circle containing the sector is the location of the weapon. We are interested in developing a fire-control system that calculates firing-solutions automatically.
The following example gives an example of a firing solution:
Figure 1

Here the firing region is the sector "ABC" that covers six points: A, B, C, D, E, H. You may further assume that the weapon is always located at point (0, 0), no targets will be on the point (0, 0) and the coordinates of the targets will be distinct.
A firing solution is called effective if and only if it covers a minimum of K points
out of N given points (targets) on the two-dimensional Cartesian plane. Furthermore,since the cost of a particular fire solution is in direct proportion to the size of the area it covers, a firing could be quite costly; thus we are only interested in the optimal firing solution with the minimum cost.
 
Input
There are multiple test cases in the input file.
Each test case starts with two non-negative integers, N and K
(1 ≤ N ≤ 5000 , K ≤ N ), followed by N lines each containing two integers, X, and Y, describing the distinct location of one target. It is guaranteed that the absolute value of any integer does not exceed 1000.
Two successive test cases are separated by a blank line. A case with N = 0 and K = 0 indicates the end of the input file, and should not be processed by your program.
 
Output
For each test case, please print the required size (to two decimal places), in the
format as indicated in the sample output.
 
Sample Input
 
3 1
0 1
1 0
-5 -6
3 2
0 2
2 0
-5 -6
0 0
 
Sample Output
 
Case #1: 0.00
Case #2: 3.14

题意:

给n个点,找出一个圆心在(0,0)的扇形,至少覆盖k个点,使得扇形的面积最小;

思路:

枚举半径r,找到与原点距离小于等于r的点,这些点极角排序后看覆盖k个点的面积,更新最小值就可以,在k==0的地方wa了几发;

AC代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map> using namespace std; #define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss)); typedef long long LL; template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + '0');
putchar('\n');
} const LL mod=998244353;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=1e4+10;
const int maxn=1e3+10;
const double eps=1e-4; struct node
{
int x,y;
double dis,ang;
}po[N];
int temp[N];
int cmp(node a,node b)
{
if(a.ang<b.ang)return 1;
return 0;
}
int main()
{
int Case=0;
while(1)
{
int n,k;
read(n);read(k);
if(!n&&!k)break;
printf("Case #%d: ",++Case);
For(i,1,n)
{
read(po[i].x),read(po[i].y);
po[i].ang=atan2(po[i].y,po[i].x);
po[i].dis=sqrt(po[i].x*po[i].x+po[i].y*po[i].y);
}
sort(po+1,po+n+1,cmp);
For(i,1,n)
{
po[i+n]=po[i];
po[i+n].ang+=2*PI;
}
double ans=1e18;
if(k==0)ans=0;
For(i,1,n)
{
double r=po[i].dis;
int cnt=0;
For(j,1,2*n)if(po[j].dis<=po[i].dis+eps)temp[++cnt]=j;
For(j,1,cnt)
{
if(temp[j]>n||j+k-1>cnt||temp[j+k-1]-temp[j]>=n)continue;
double angle=po[temp[j+k-1]].ang-po[temp[j]].ang;
ans=min(ans,angle*r*r/2);
}
}
printf("%.2lf\n",ans);
}
return 0;
}

  

LA-4356&&hdu-2469 (极角排序+扫描线)的更多相关文章

  1. POJ 2280 Amphiphilic Carbon Molecules 极角排序 + 扫描线

    从TLE的暴力枚举 到 13313MS的扫描线  再到 1297MS的简化后的扫描线,简直感觉要爽翻啦.然后满怀欣喜的去HDU交了一下,直接又回到了TLE.....泪流满面 虽说HDU的时限是2000 ...

  2. 【极角排序、扫描线】UVa 1606 - Amphiphilic Carbon Molecules(两亲性分子)

    Shanghai Hypercomputers, the world's largest computer chip manufacturer, has invented a new class of ...

  3. HDU 5738 Eureka(极角排序)

    [题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=5738 [题目大意] 给出平面中一些点,在同一直线的点可以划分为一个集合,问可以组成多少包含元素不少 ...

  4. 【极角排序+双指针线性扫】2017多校训练七 HDU 6127 Hard challenge

    acm.hdu.edu.cn/showproblem.php?pid=6127 [题意] 给定平面直角坐标系中的n个点,这n个点每个点都有一个点权 这n个点两两可以连乘一条线段,定义每条线段的权值为线 ...

  5. HDU Always Cook Mushroom (极角排序+树状数组)

    Problem Description Matt has a company, Always Cook Mushroom (ACM), which produces high-quality mush ...

  6. 2017ACM暑期多校联合训练 - Team 7 1008 HDU 6127 Hard challenge (极角排序)

    题目链接 Problem Description There are n points on the plane, and the ith points has a value vali, and i ...

  7. 【极角排序】【扫描线】hdu6127 Hard challenge

    平面上n个点,每个点带权,任意两点间都有连线,连线的权值为两端点权值之积.没有两点连线过原点.让你画一条过原点直线,把平面分成两部分,使得直线穿过的连线的权值和最大. 就把点极角排序后,扫过去,一侧的 ...

  8. LA 4064 (计数 极角排序) Magnetic Train Tracks

    这个题和UVa11529很相似. 枚举一个中心点,然后按极角排序,统计以这个点为钝角的三角形的个数,然后用C(n, 3)减去就是答案. 另外遇到直角三角形的情况很是蛋疼,可以用一个eps,不嫌麻烦的话 ...

  9. poj2280--Amphiphilic Carbon Molecules(扫描线+极角排序+转换坐标)

    题目链接:id=2280">点击打开链接 题目大意:给出n个点的坐标.每一个点有一个值0或者1,如今有一个隔板(无限长)去分开着n个点,一側统计0的个数,一側统计1的个数,假设点在板上 ...

随机推荐

  1. hanzi 全拼音 qu de

    Function pinyin(ByVal mystr As String, Optional types As Byte = 0) As StringDim temp   As String, i ...

  2. T1230 元素查找 codevs

    http://codevs.cn/problem/1230/  题目描述 Description 给出n个正整数,然后有m个询问,每个询问一个整数,询问该整数是否在n个正整数中出现过. 输入描述 In ...

  3. UVA 11346 Probability

    题目描述 PDF 输入输出格式 输入格式: 输出格式: 输入输出样例 输入样例#1: 3 10 5 20 1 1 1 2 2 0 输出样例#1: 23.348371% 0.000000% 100.00 ...

  4. tensorflow基础练习:线性模型

    TensorFlow是一个面向数值计算的通用平台,可以方便地训练线性模型.下面采用TensorFlow完成Andrew Ng主讲的Deep Learning课程练习题,提供了整套源码. 线性回归 多元 ...

  5. BUPT复试专题—最长连续等差子数列(2014软院)

    题目描述   给定-个长度为N的整数数列,你需要在其中找到最长的连续子数列的长度, 并满足这个子数列是等差的.注意公差小于或等于0的情况也是允许的. 输入 第一行为数据组数T(1~100),表示测试数 ...

  6. BUPT复试专题—求导数(2015)

    题目描述 描述:求函数f(x) = a*x^3 + b*x^2 + c*x + d在x = x0处的一阶导数.   输入 数据第一行是数据的组数m 接下来m行的每一行分别是 a b c d x0 输出 ...

  7. BUPT复试专题—二进制数(2012)

    https://www.nowcoder.com/practice/103dd589fed14457a673c613d8de3841?tpId=67&tqId=29634&tPage= ...

  8. onlyOffice 开发相关 总结

    onlyOffice 服务端 客户端 相关开发整理 功能: 所有客户端都可用 云端部署服务 查看 预览 doc ppt excel 编辑 权限控制 开发技术准备 用户服务器端 提供保存接口 用户浏览器 ...

  9. row and col

    1.行 <Row gutter={{ md: 6, lg: 12, xl: 12 }}></Row> gutter: md: 中等屏幕 桌面显示器 (≥992px) lg: 大 ...

  10. Linux下的ELF可执行文件的格式解析 (转)

    LInux命令只是和Kernel一起被编译进操作系统的存在于FS的ELF格式二进制文件,或者权限足够的脚本,或者一个软链 ELF(Executable and Linking Format)是一种对象 ...