hdu 2295 Radar (二分答案+重复覆盖)

题目来源：http://acm.hdu.edu.cn/showproblem.php?pid=2295

Radar

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1023    Accepted Submission(s): 405

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

题目大意：一个国家有n个城市，有m个地方可以建造雷达，最多可以建K个雷达（K>=1 && K<=m）,问雷达最短的探测半径，才能使n个城市都能探测到。

思路：比较裸一点的Dancing Links，二分答案，然后算出当前状况下 重复覆盖所需的雷达的个数，判断能否满足条件。

对于这样的题 精度是比较恶心的，，我开始用1e-7，无论怎样二分都是wa，改成1e-8之后就能AC。。不过有的就是用1e-7过的。ORZ。。

4355580	2011-08-07 18:35:34	Accepted	2295	171MS	228K	2134 B	G++	zyzamp
4355579	2011-08-07 18:35:15	Wrong Answer	2295	171MS	228K	2134 B	G++	zyzamp
4355567	2011-08-07 18:32:46	Wrong Answer	2295	156MS	228K	2139 B	G++	zyzamp
4355553	2011-08-07 18:29:21	Accepted	2295	171MS	228K	2120 B	G++	zyzamp
4355527	2011-08-07 18:23:49	Wrong Answer	2295	31MS	228K	2055 B	G++	zyzamp
4355520	2011-08-07 18:21:54	Accepted	2295	375MS	228K	2084 B	G++	zyzamp
4355500	2011-08-07 18:13:43	Wrong Answer	2295	437MS	228K	2090 B	G++	zyzamp
4355498	2011-08-07 18:13:14	Wrong Answer	2295	406MS	228K	2090 B	G++	zyzamp
4355497	2011-08-07 18:13:03	Compilation Error	2295	0MS	0K	2090 B	C++	zyzamp

code：

View Code

  1 Problem : 2295 ( Radar )     Judge Status : Accepted
  2 RunId : 4355553    Language : G++    Author : zhuyawei
  3 Code Render Status : Rendered By HDOJ G++ Code Render Version 0.01 Beta
  4 # include<stdio.h>
  5 # include<math.h>
  6 # include<string.h>
  7 # define eps 1e-8
  8 # define N 55
  9 # define V 3600
 10 int n,m,K;
 11 int L[V],R[V];
 12 int D[V],U[V];
 13 int C[V];
 14 int S[N],H[N];
 15 int ak,size;
 16 double dis(double x1,double y1,double x2,double y2)
 17 {
 18     return sqrt((x2-x1)*(x2-x1) + (y2-y1)*(y2-y1));
 19 }
 20 void Link(int r,int c)
 21 {
 22     S[c]++;C[size]=c;
 23     U[size]=U[c];D[U[c]]=size;
 24     D[size]=c;U[c]=size;
 25     if(H[r]==-1) H[r]=L[size]=R[size]=size;
 26     else
 27     {
 28         L[size]=L[H[r]];R[L[H[r]]]=size;
 29         R[size]=H[r];L[H[r]]=size;
 30     }
 31     size++;
 32 }
 33 void remove(int c)
 34 {
 35     int i;
 36     for(i=D[c];i!=c;i=D[i])
 37         L[R[i]]=L[i],R[L[i]]=R[i];
 38 }
 39 void resume(int c)
 40 {
 41     int i;
 42     for(i=U[c];i!=c;i=U[i])
 43         L[R[i]]=R[L[i]]=i;
 44 }
 45 int h()
 46 {
 47     int i,j,k,count=0;
 48     bool visit[N];
 49     memset(visit,0,sizeof(visit));
 50     for(i=R[0];i;i=R[i])
 51     {
 52         if(visit[i]) continue;
 53         count++;
 54         visit[i]=1;
 55         for(j=D[i];j!=i;j=D[j])
 56         {
 57             for(k=R[j];k!=j;k=R[k])
 58                 visit[C[k]]=1;
 59         }
 60     }
 61     return count;
 62 }
 63 void Dance(int k)
 64 {
 65     int i,j,c,Min,ans;
 66     ans=h();
 67     if(k+ans>K || k+ans>=ak) return;
 68     if(!R[0])
 69     {
 70         if(k<ak) ak=k;
 71         return;
 72     }
 73     for(Min=N,i=R[0];i;i=R[i])
 74         if(S[i]<Min) Min=S[i],c=i;
 75     for(i=D[c];i!=c;i=D[i])
 76     {
 77         remove(i);
 78         for(j=R[i];j!=i;j=R[j])
 79             remove(j);
 80         Dance(k+1);
 81         for(j=L[i];j!=i;j=L[j])
 82             resume(j);
 83         resume(i);
 84     }
 85     return;
 86 }
 87 int main()
 88 {
 89     int i,j,ncase;
 90     double x[N],y[N],x1[N],y1[N];
 91     double left,right,ans,mid;
 92     scanf("%d",&ncase);
 93     while(ncase--)
 94     {
 95         scanf("%d%d%d",&n,&m,&K);
 96         for(i=1;i<=n;i++)
 97             scanf("%lf%lf",&x[i],&y[i]);
 98         for(i=1;i<=m;i++)
 99             scanf("%lf%lf",&x1[i],&y1[i]);
100         left=0;
101         right=1416.0;
102         ans=right;
103         while(right>=left)
104         {
105             for(i=0;i<=n;i++)
106             {
107                 S[i]=0;
108                 U[i]=D[i]=i;
109                 L[i+1]=i;R[i]=i+1;
110             }R[n]=0;
111             memset(H,-1,sizeof(H));
112             size=n+1;
113             mid=(left+right)/2;
114             for(i=1;i<=m;i++)
115             {
116                 for(j=1;j<=n;j++)
117                     if(mid>=dis(x1[i],y1[i],x[j],y[j]))  Link(i,j); 
118             }
119             ak=N;
120             Dance(0);
121             if(ak<=K) {ans=mid<ans?mid:ans;right=mid-eps;}
122             else left=mid+eps;
123         }
124         printf("%.6lf\n",ans);
125     }
126     return 0;
127 }