hdu 2295 Radar（Dancing Links重复覆盖）

Radar

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2903 Accepted Submission(s): 1124



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

n个城市 m个雷达 现在最多使用K个雷达覆盖所有的城市 求出雷达覆盖半径的最小值

二分距离 然后求出当前情况下 使用雷达最少的数量 如果符合要求 那就缩小当前距离 如果不符合就扩大

DancingLinks 感觉重复覆盖建图 比精确覆盖要难。。。

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>
#include <vector>
#include <queue>

#define MEM(a,x) memset(a,x,sizeof a)
#define eps 1e-8
#define MOD 10009
#define INF 99999999
#define ll __int64
#define bug cout<<"here"<<endl
#define fread freopen("ceshi.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)

using namespace std;

const int MAXM=60;
const int MAXN=60;
const int maxnode=MAXN*MAXM;

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[MAXM];
    int ansd;
    void init(int _n,int _m)
    {
        n=_n; m=_m;
        for(int i=0;i<=m;i++)
        {
            S[i]=0;
            U[i]=D[i]=i;
            L[i]=i-1;
            R[i]=i+1;
        }
        R[m]=0; L[0]=m;
        size=m;
        for(int i=1;i<=n;i++) H[i]=-1;
    }
    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]<0) 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[MAXM];
    int f()
    {
        int ret=0;
        for(int c=R[0];c!=0;c=R[c]) v[c]=1;
        for(int c=R[0];c!=0;c=R[c])
        {
            if(v[c])
            {
                ret++;
                v[c]=0;
                for(int i=D[c];i!=c;i=D[i])
                    for(int j=R[i];j!=i;j=R[j])
                        v[Col[j]]=0;
            }
        }
        return ret;
    }
    bool Dance(int d)
    {
//        cout<<d<<endl;
        if(d+f()>K) return 0;//个数大于K 不符合要求
        if(R[0]==0)
        {
            if(d<=K) return 1;
            else return 0;
        }
        int c=R[0];
        for(int i=R[0];i!=0;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+1)) return 1;
            for(int j=L[i];j!=i;j=L[j]) resume(j);
            resume(i);
        }
        return 0;
    }
};
DLX dlx;
int x[60],y[60],xx[60],yy[60];
double dis(int a,int b,int c,int d)
{
    double len=(double)((c-a)*(c-a)+(d-b)*(d-b));
    return sqrt(len);
}

int main()
{
//    fread;
    int tc;
    scanf("%d",&tc);
    while(tc--)
    {
        int n,m;
        scanf("%d%d%d",&n,&m,&K);
        for(int i=0;i<n;i++)
            scanf("%d%d",&x[i],&y[i]);
        for(int i=0;i<m;i++)
            scanf("%d%d",&xx[i],&yy[i]);
        double left=0.0,right=1500.0,mid;
        while(right-left>eps)
        {
//            bug;
            dlx.init(m,n);
            mid=(left+right)/2.0;
            for(int i=0;i<m;i++)
                for(int j=0;j<n;j++)
            {
                if(mid-eps>(dis(xx[i],yy[i],x[j],y[j])))
                    dlx.Link(i+1,j+1);
            }
            if(dlx.Dance(0)) right=mid-eps;
            else left=mid+eps;
        }
        printf("%.6lf\n",right);
    }
    return 0;
}