Radar

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2280    Accepted Submission(s): 897

Problem Description
N cities of the Java Kingdom need to be covered by radars for being in a state of war. Since the kingdom has M radar stations but only K operators, we can at most operate K radars. All radars have the same circular coverage with a radius of R. Our goal is to minimize R while covering the entire city with no more than K radars.
 
Input
The input consists of several test cases. The first line of the input consists of an integer T, indicating the number of test cases. The first line of each test case consists of 3 integers: N, M, K, representing the number of cities, the number of radar stations and the number of operators. Each of the following N lines consists of the coordinate of a city.
Each of the last M lines consists of the coordinate of a radar station.

All coordinates are separated by one space.
Technical Specification

1. 1 ≤ T ≤ 20
2. 1 ≤ N, M ≤ 50
3. 1 ≤ K ≤ M
4. 0 ≤ X, Y ≤ 1000

 
Output
For each test case, output the radius on a single line, rounded to six fractional digits.
 
Sample Input
1
3 3 2
3 4
3 1
5 4
1 1
2 2
3 3
 
Sample Output
2.236068
 
Source
 

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2295

二分答案, 然后使用重复覆盖的Dancing Links模板进行判断,看使用K个能不能覆盖n个点

 /* ***********************************************
Author :kuangbin
Created Time :2014/5/26 22:20:05
File Name :E:\2014ACM\专题学习\DLX\HDU2295.cpp
************************************************ */ #include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const int maxnode = ;
const int MaxM = ;
const int MaxN = ;
int K;
struct DLX
{
int n,m,size;
int U[maxnode],D[maxnode],R[maxnode],L[maxnode],Row[maxnode],Col[maxnode];
int H[MaxN],S[MaxN];
int ands,ans[MaxN];
void init(int _n,int _m)
{
n = _n;
m = _m;
for(int i = ;i <= m;i++)
{
S[i] = ;
U[i] = D[i] = i;
L[i] = i-;
R[i] = i+;
}
R[m] = ; L[] = m;
size = m;
for(int i = ;i <= n;i++)
H[i] = -;
}
void Link(int r,int c)
{
++S[Col[++size]=c];
Row[size] = r;
D[size] = D[c];
U[D[c]] = size;
U[size] = c;
D[c] = size;
if(H[r] < )H[r] = L[size] = R[size] = size;
else
{
R[size] = R[H[r]];
L[R[H[r]]] = size;
L[size] = H[r];
R[H[r]] = size;
}
}
void remove(int c)
{
for(int i = D[c];i != c;i = D[i])
L[R[i]] = L[i], R[L[i]] = R[i];
}
void resume(int c)
{
for(int i = U[c];i != c;i = U[i])
L[R[i]]=R[L[i]]=i;
}
bool v[maxnode];
int f()
{
int ret = ;
for(int c = R[];c != ;c = R[c])v[c] = true;
for(int c = R[];c != ;c = R[c])
if(v[c])
{
ret++;
v[c] = false;
for(int i = D[c];i != c;i = D[i])
for(int j = R[i];j != i;j = R[j])
v[Col[j]] = false;
}
return ret; }
bool Dance(int d)
{
if(d + f() > K)return false;
if(R[] == )return d <= K;
int c = R[];
for(int i = R[];i != ;i = R[i])
if(S[i] < S[c])
c = i;
for(int i = D[c];i != c;i = D[i])
{
remove(i);
for(int j = R[i];j != i;j = R[j])remove(j);
if(Dance(d+))return true;
for(int j = L[i];j != i;j = L[j])resume(j);
resume(i);
}
return false;
}
};
DLX g;
const double eps = 1e-;
struct Point
{
int x,y;
void input()
{
scanf("%d%d",&x,&y);
}
}city[MaxM],station[MaxN];
double dis(Point a,Point b)
{
return sqrt((double)(a.x-b.x)*(a.x-b.x)+(double)(a.y-b.y)*(a.y-b.y));
} int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int T;
int n,m;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&n,&m,&K);
for(int i = ;i < n;i++)city[i].input();
for(int i = ;i < m;i++)station[i].input();
double l = , r = 1e8;
while(r-l >= eps)
{
double mid = (l+r)/;
g.init(m,n);
for(int i = ;i < m;i++)
for(int j = ;j < n;j++)
if(dis(station[i],city[j]) < mid - eps)
g.Link(i+,j+);
if(g.Dance())r = mid-eps;
else l = mid+eps;
}
printf("%.6lf\n",l);
}
return ;
}

HDU 2295 Radar (重复覆盖)的更多相关文章

  1. HDU 2295 Radar 重复覆盖 DLX

    题意: N个城市,M个雷达站,K个操作员,问雷达的半径至少为多大,才能覆盖所有城市.M个雷达中最多只能有K个同时工作. 思路: 二分雷达的半径,看每个雷达可以覆盖哪些城市,然后做重复覆盖,判断这个半径 ...

  2. hdu 2295 Radar 重复覆盖+二分

    题目链接 给m个雷达, n个城市, 以及每个城市的坐标, m个雷达里只能使用k个, 在k个雷达包围所有城市的前提下, 求最小半径. 先求出每个雷达到所有城市的距离, 然后二分半径, 如果距离小于二分的 ...

  3. hdu 2295 dlx重复覆盖+二分答案

    题目大意: 有一堆雷达工作站,安放至多k个人在这些工作站中,找到一个最小的雷达监控半径可以使k个工作人所在的雷达工作站覆盖所有城市 二分半径的答案,每次利用dlx的重复覆盖来判断这个答案是否正确 #i ...

  4. (中等) HDU 2295 , DLX+重复覆盖+二分。

    Description N cities of the Java Kingdom need to be covered by radars for being in a state of war. S ...

  5. HDU 2295 Radar (二分 + Dancing Links 重复覆盖模型 )

    以下转自 这里 : 最小支配集问题:二分枚举最小距离,判断可行性.可行性即重复覆盖模型,DLX解之. A*的启发函数: 对当前矩阵来说,选择一个未被控制的列,很明显该列最少需要1个行来控制,所以ans ...

  6. HDU 2295 Radar dancing links 重复覆盖

    就是dancing links 求最小支配集,重复覆盖 精确覆盖时:每次缓存数据的时候,既删除行又删除列(这里的删除列,只是删除表头) 重复覆盖的时候:只删除列,因为可以重复覆盖 然后重复覆盖有一个估 ...

  7. HDU 2295.Radar (DLX重复覆盖)

    2分答案+DLX判断可行 不使用的估计函数的可重复覆盖的搜索树将十分庞大 #include <iostream> #include <cstring> #include < ...

  8. hdu 3498 whosyourdaddy 重复覆盖

    题目链接 重复覆盖的入门题, 和精确覆盖不一样, 删除的时候只删除一行多列. #include<bits/stdc++.h> using namespace std; #define pb ...

  9. (中等) HDU 3335 , DLX+重复覆盖。

    Description As we know,the fzu AekdyCoin is famous of math,especially in the field of number theory. ...

随机推荐

  1. 字典树的C++实现

    此次代码使用了诸多新鲜玩意,比如自动类型推断,模板类等等,感觉真的超级好玩啊. 自己做了一个简易的测试,程序的健壮性什么的还是比较差的,此程序留待以后进行功能补全. #pragma once #inc ...

  2. 【转载】快速收索并更新sid 方法

    利用Google的搜索功能,可以获得不少SAS各个版本的SID号,试过之后你会异常惊喜.1.打开谷歌: http://google.com.hk2.输入或复制这个段文字:"SID_heade ...

  3. 思考方式--SMART原则

    如果你对项目管理.系统架构有兴趣,请加微信订阅号"softjg",加入这个PM.架构师的大家庭 万事开头于你目标的设定,如果开始走错了,那么后面的路将会更加的错误,甚至于更加的努力 ...

  4. 冒泡排序,sql分页语句

    对数组中的数字进行排序 public int[] PopSmall(int[] IntArray) { ; ; i < IntArray.Length - ; i++) { ; j < I ...

  5. ACM知识点

    基础算法 高精 模拟 分治 贪心 排序 DFS 迭代加深搜索 BFS 双向BFS 动态规划 DAG上DP 树上DP 线性DP 图算法 最短路 FLYD DJATL BF 最大流 Dinic ISAP ...

  6. 【原创】安装LoadRunner12.53 版本时出现Critical error的解决方法

    步骤: 1.在官网上下载LoadRunner12.53正版,只不过要注册,然后官网会给个序列号. 2.安装成功之后,快捷键已创建,打开Virtual User Generator时,提示如下错误: 此 ...

  7. Salesforce 使用Js 调用Webservice实例

    1,创建 Custom Button 在页面上 2, 创建CustomJs 代码调用Webservice <!--参数名区分大小写,对于跨层object直接在Object名后直接加参字段名即可- ...

  8. C# List结果集排序

    public class Student { public string Name { get; set; } public int Age { get; set; } public Student( ...

  9. PowerDesigner 把Comment复制到name中和把name复制到Comment

    在使用PowerDesigner对数据库进行概念模型和物理模型设计时,一般在NAME或Comment中写中文,在Code中写英文.Name用来显 示,Code在代码中使用,但Comment中的文字会保 ...

  10. 答:SQLServer DBA 三十问之五:有关视图索引

    5. 视图上我们能建索引吗,如果能建的话,会啥好处和坏处:视图上建索引和表上建索引有啥区别: ----------------------------------------------------- ...