(中等) HDU 2295 , DLX+重复覆盖+二分。

　　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.

 

　　题目就是让我们找到一个最小的R,能够在M个站里面找到K个来覆盖N个城市。

 

　　假设知道R的话，就可以通过DLX判断这个R成不成立，然后二分查找R就好。。。。。。

 

代码如下：

#include<iostream>
#include<cstring>

using namespace std;

const int MaxN=55;
const int MaxM=55;
const int MaxNode=MaxN*MaxM;

int K;
int N,M;

struct DLX
{
    int D[MaxNode],U[MaxNode],L[MaxNode],R[MaxNode],col[MaxNode];
    int H[MaxN],S[MaxM];
    int n,m,size;

    void init(int _n,int _m)
    {
        n=_n;
        m=_m;

        for(int i=0;i<=m;++i)
        {
            D[i]=U[i]=i;
            L[i]=i-1;
            R[i]=i+1;
            
            S[i]=0;
        }
        L[0]=m;
        R[m]=0;
        
        size=m;

        for(int i=0;i<=n;++i)
            H[i]=-1;
    }

    void Link(int r,int c)
    {
        col[++size]=c;
        ++S[c];

        U[size]=U[c];
        D[size]=c;
        D[U[c]]=size;
        U[c]=size;

        if(H[r]==-1)
            H[r]=L[size]=R[size]=size;
        else
        {
            L[size]=L[H[r]];
            R[size]=H[r];
            R[L[H[r]]]=size;
            L[H[r]]=size;
        }
    }

    void remove(int c)
    {
        for(int i=D[c];i!=c;i=D[i])
        {
            R[L[i]]=R[i];
            L[R[i]]=L[i];
        }
    }

    void resume(int c)
    {
        for(int i=U[c];i!=c;i=U[i])
            R[L[i]]=L[R[i]]=i;
    }

    bool vis[MaxM];

    int getH()
    {
        int ret=0;

        for(int c=R[0];c!=0;c=R[c])
            vis[c]=1;

        for(int c=R[0];c!=0;c=R[c])
            if(vis[c])
            {
                ++ret;
                vis[c]=0;

                for(int i=D[c];i!=c;i=D[i])
                    for(int j=R[i];j!=i;j=R[j])
                        vis[col[j]]=0;
            }

        return ret;
    }

    bool Dance(int d)
    {
        if(d+getH()>K)
            return 0;

        if(R[0]==0)
        {
            if(d<=K)
                return 1;
            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 Rx[60],Ry[60],Cx[60],Cy[60];

void solve()
{
    double L=0,R=1001.0,Mid;

    while(R-L>0.0000001)
    {
        Mid=(L+R)/2;

        dlx.init(M,N);

        for(int i=1;i<=M;++i)
            for(int j=1;j<=N;++j)
                if(Mid*Mid>=(Rx[i]-Cx[j])*(Rx[i]-Cx[j])+(Ry[i]-Cy[j])*(Ry[i]-Cy[j]))
                    dlx.Link(i,j);

        if(dlx.Dance(0))
            R=Mid;
        else
            L=Mid;
    }

    cout<<L<<endl;
}

int main()
{
    ios::sync_with_stdio(false);
    cout.setf(ios::fixed);
    cout.precision(6);

    int T;
    cin>>T;

    while(T--)
    {
        cin>>N>>M>>K;

        for(int i=1;i<=N;++i)
            cin>>Cx[i]>>Cy[i];

        for(int i=1;i<=M;++i)
            cin>>Rx[i]>>Ry[i];

        solve();
    }

    return 0;
}