HDU 2295 DLX 二分

Radar

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2229    Accepted Submission(s): 888

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

13 3 23 43 15 41 12 23 3

 

Sample Output

2.236068

有n个城市，m个发电站，已知他们的坐标，每个发电站覆盖半径相同，求多大半径时可以把所有的城市都覆盖，二分半径建图，dlx爆搜。

代码：

/* ***********************************************Author :_rabbitCreated Time :2014/4/28 14:08:33File Name :1.cpp************************************************ */#pragma comment(linker, "/STACK:102400000,102400000")#include <stdio.h>#include <iostream>#include <algorithm>#include <sstream>#include <stdlib.h>#include <string.h>#include <limits.h>#include <string>#include <time.h>#include <math.h>#include <queue>#include <stack>#include <set>#include <map>using namespace std;#define INF 0x3f3f3f3f#define eps 1e-8#define pi acos(-1.0)typedef long long ll;struct Point{    double x,y;}city[100],radar[100];double dis[66][66];double dist(Point a,Point b){    double s=a.x-b.x,t=a.y-b.y;    return s*s+t*t;}int K;struct DLX{    const static int maxn=20010;    #define FF(i,A,s) for(int i = A[s];i != s;i = A[i])    int L[maxn],R[maxn],U[maxn],D[maxn];    int size,col[maxn],row[maxn],s[maxn],H[maxn];    bool vis[70];    int ans[maxn],cnt;    void init(int m){        for(int i=0;i<=m;i++){            L[i]=i-1;R[i]=i+1;U[i]=D[i]=i;s[i]=0;        }        memset(H,-1,sizeof(H));        L[0]=m;R[m]=0;size=m+1;    }    void link(int r,int c){         U[size]=c;D[size]=D[c];U[D[c]]=size;D[c]=size;         if(H[r]<0)H[r]=L[size]=R[size]=size;         else {             L[size]=H[r];R[size]=R[H[r]];             L[R[H[r]]]=size;R[H[r]]=size;         }         s[c]++;col[size]=c;row[size]=r;size++;     }    void del(int c){//精确覆盖        L[R[c]]=L[c];R[L[c]]=R[c];          FF(i,D,c)FF(j,R,i)U[D[j]]=U[j],D[U[j]]=D[j],--s[col[j]];      }      void add(int c){  //精确覆盖        R[L[c]]=L[R[c]]=c;          FF(i,U,c)FF(j,L,i)++s[col[U[D[j]]=D[U[j]]=j]];      }      bool dfs(int k){//精确覆盖        if(!R[0]){              cnt=k;return 1;          }          int c=R[0];FF(i,R,0)if(s[c]>s[i])c=i;          del(c);          FF(i,D,c){              FF(j,R,i)del(col[j]);              ans[k]=row[i];if(dfs(k+1))return true;              FF(j,L,i)add(col[j]);          }          add(c);          return 0;    }      void remove(int c){//重复覆盖        FF(i,D,c)L[R[i]]=L[i],R[L[i]]=R[i];    }     void resume(int c){//重复覆盖         FF(i,U,c)L[R[i]]=R[L[i]]=i;     }    int A(){//估价函数        int res=0;        memset(vis,0,sizeof(vis));        FF(i,R,0)if(!vis[i]){                res++;vis[i]=1;                FF(j,D,i)FF(k,R,j)vis[col[k]]=1;            }        return res;    }    void dfs(int now,int &ans){//重复覆盖        if(R[0]==0)ans=min(ans,now);        else if(now+A()<ans){            int temp=INF,c;            FF(i,R,0)if(temp>s[i])temp=s[i],c=i;            FF(i,D,c){                remove(i);FF(j,R,i)remove(j);                dfs(now+1,ans);                FF(j,L,i)resume(j);resume(i);            }        }    }/*    int astar(){        int lim=A();        while(!dfs(0,lim))lim++;        return lim;    }*/}dlx;int main(){     //freopen("data.in","r",stdin);     //freopen("data.out","w",stdout);     int T,n,m;     cin>>T;     while(T--){         scanf("%d%d%d",&n,&m,&K);         for(int i=1;i<=n;i++)scanf("%lf%lf",&city[i].x,&city[i].y);         for(int i=1;i<=m;i++)scanf("%lf%lf",&radar[i].x,&radar[i].y);         for(int i=1;i<=m;i++)             for(int j=1;j<=n;j++)dis[i][j]=dist(radar[i],city[j]);         double left=0,right=1500;         while(right-left>1e-7){             double mid=(left+right)/2;             double ss=mid*mid;             dlx.init(n);             for(int i=1;i<=m;i++)                 for(int j=1;j<=n;j++)                     if(dis[i][j]<=ss)dlx.link(i,j);        //     cout<<left<<" "<<right<<" "<<ans<<" "<<dlx.size<<endl;             int ans=INF;             dlx.dfs(0,ans);             if(ans>K)left=mid;             else right=mid;         }         printf("%.6lf\n",left);     }     return 0;}