zoukankan      html  css  js  c++  java
  • 搜索(DLX重复覆盖模板):HDU 2295 Radar

     

    Radar

     

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 3684    Accepted Submission(s): 1398


    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
      水题,主要是为了贴模板。
      1 #include <iostream>
      2 #include <cstring>
      3 #include <cstdio>
      4 #include <cmath>
      5 using namespace std;
      6 const double eps=1e-8;
      7 const int N=55,M=3005;
      8 struct Point{int x,y;}c[N],r[N];
      9 double sqr(double x){return 1.0*x*x;}
     10 double dis(Point a,Point b){return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));}
     11 
     12 void P(int x){
     13   printf("%d
    ",x);
     14 }
     15 
     16 int T,n,m,k,col[M],row[M];
     17 int U[M],D[M],L[M],R[M];
     18 int H[N],C[N],vis[N],cnt;
     19 struct DLX{
     20   void Init(int n,int m){
     21     for(int i=0;i<=m;i++){
     22       L[i]=i-1;R[i]=i+1;
     23       C[i]=0;U[i]=D[i]=i;
     24     }L[0]=m;R[m]=0;cnt=m;
     25     for(int i=1;i<=n;i++)H[i]=0;
     26   }
     27   void Link(int r,int c){
     28     C[c]+=1;++cnt;
     29     U[D[c]]=cnt;U[cnt]=c;
     30     D[cnt]=D[c];D[c]=cnt;
     31     row[cnt]=r;col[cnt]=c;
     32     
     33     if(H[r]){
     34       L[R[H[r]]]=cnt;L[cnt]=H[r];
     35       R[cnt]=R[H[r]];R[H[r]]=cnt;
     36     }
     37     else H[r]=L[cnt]=R[cnt]=cnt;
     38   }
     39   void Delete(int x){
     40     for(int i=D[x];i!=x;i=D[i])
     41       L[R[i]]=L[i],R[L[i]]=R[i];
     42   }
     43   void Resume(int x){
     44     for(int i=U[x];i!=x;i=U[i])
     45       L[R[i]]=i,R[L[i]]=i;
     46   }
     47   int F(){
     48     int ret=0;
     49     for(int c=R[0];c;c=R[c])vis[c]=0;
     50     for(int c=R[0];c;c=R[c]){
     51       if(vis[c])continue;ret+=1;
     52       for(int i=D[c];i!=c;i=D[i])
     53     for(int j=R[i];j!=i;j=R[j])
     54       vis[col[j]]=1;vis[c]=1;
     55     }
     56     return ret;
     57   }
     58   bool Dance(int dep){
     59     if(!R[0])return dep<=k;
     60     if(dep+F()>k)return false;
     61     int p=0;
     62     for(int i=R[0];i;i=R[i])
     63       if(!p||C[i]<C[p])p=i;
     64     for(int i=D[p];i!=p;i=D[i]){
     65       Delete(i);
     66       for(int j=R[i];j!=i;j=R[j])Delete(j);
     67       if(Dance(dep+1))return true;
     68       for(int j=L[i];j!=i;j=L[j])Resume(j);
     69       Resume(i);
     70     }
     71     return false;
     72   }
     73 
     74   bool Check(double d){
     75     Init(m,n);
     76     for(int i=1;i<=m;i++)
     77       for(int j=1;j<=n;j++)
     78     if(d>=dis(r[i],c[j]))
     79       Link(i,j);
     80     return Dance(0);
     81   }
     82 }dlx;
     83 
     84 int main(){
     85   scanf("%d",&T);
     86   while(T--){
     87     scanf("%d%d%d",&n,&m,&k);
     88     for(int i=1;i<=n;i++)
     89       scanf("%d%d",&c[i].x,&c[i].y);
     90     for(int i=1;i<=m;i++)
     91       scanf("%d%d",&r[i].x,&r[i].y);
     92     double l=0,r=1e3;
     93     while(r-l>=eps){
     94       double mid=(l+r)/2;
     95       if(dlx.Check(mid))r=mid;
     96       else l=mid;
     97     }
     98     printf("%.6f
    ",l);
     99   }
    100   return 0;
    101 }
  • 相关阅读:
    不一样的图片加载方式
    赢 1000 元现金红包!助力奥运,猜金银牌数赢现金
    接入 SDK 结果翻车了?了解 SDK 的那些事
    关于 IPv6 国家有大动作啦!快来瞅瞅行动计划都说了什么~
    MySQL 那些常见的错误设计规范
    webpack 从 0 到 1 构建 vue
    PHP 网络通信底层原理分析
    内部方案汇总
    taro+vue3 引入 taro-ui-vue3
    springboot+tomcat+vue+nginx 前后端分离配置
  • 原文地址:https://www.cnblogs.com/TenderRun/p/6062956.html
Copyright © 2011-2022 走看看