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.
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
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 ;
}