(中等) HDU 5046 Airport ，DLX+可重复覆盖+二分。

　　Description

　　The country of jiuye composed by N cites. Each city can be viewed as a point in a two- dimensional plane with integer coordinates (x,y). The distance between city i and city j is defined by d _ij = |x _i - x _j| + |y _i - y _j|. jiuye want to setup airport in K cities among N cities. So he need your help to choose these K cities, to minimize the maximum distance to the nearest airport of each city. That is , if we define d _i(1 ≤ i ≤ N ) as the distance from city i to the nearest city with airport. Your aim is to minimize the value max{d _i|1 ≤ i ≤ N }. You just output the minimum.

　　题目是最大值最小化，明显的二分，然后对于每一个值，判断能不能找到K个以内的来覆盖。。。。。。

　　做的时候要注意二分的问题，还有就是要用long long 别用int，还有就是abs。。。不知道怎么回事被坑了，自己写了一个abs来用的。。。。。。

#include<iostream>
#include<cstring>
#include<utility>
#include<algorithm>

using namespace std;

const int MaxN=70;
const int MaxM=70;
const int MaxNode=MaxN*MaxM;

int N,K;
long long X[70],Y[70];

long long abs1(long long x)
{
    if(x<0)
        x=-x;

    return x;
}

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

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

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

            S[i]=0;
        }

        L[0]=m;
        R[m]=0;

        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])
        {
            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 vis[MaxM];

    int getH()
    {
        int ret=0;

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

        for(int c=R[0];c;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)
            return d<=K;

        int c=R[0];

        for(int i=R[0];i;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;
    }
};

pair <long long,int> rem[70][70];
long long remP[70];
DLX dlx;

void slove(long long maxNum)
{
    long long L=0,R=maxNum,M,ans;

    for(int i=1;i<=N;++i)
    {
        for(int j=1;j<=N;++j)
        {
            rem[i][j].first=(long long)abs1(X[i]-X[j])+(long long)abs1(Y[i]-Y[j]);
            rem[i][j].second=j;
        }

        sort(rem[i]+1,rem[i]+N+1);

        remP[i]=1;
    }

/*    dlx.init(N,N);

    for(M=L;M<=R;++M)
    {
        for(int i=1;i<=N;++i)
            while(rem[i][remP[i]].first<=M)
            {
                dlx.Link(i,rem[i][remP[i]].second);
                ++remP[i];
            }

        if(dlx.Dance(0))
            break;
    }*/

    while(R>L)
    {
        M=(L+R)/2;

        dlx.init(N,N);

        for(int i=1;i<=N;++i)
            for(int j=1;j<=N;++j)
                if(rem[i][j].first<=M)
                    dlx.Link(i,rem[i][j].second);
                else
                    break;

        if(dlx.Dance(0))
            R=M;
        else
            L=M+1;
    }

    cout<<L<<endl;
}

int main()
{
    ios::sync_with_stdio(false);

    int T;
    long long maxX,maxY,minX,minY;
    cin>>T;

    for(int cas=1;cas<=T;++cas)
    {
        cin>>N>>K;

        cin>>X[1]>>Y[1];

        maxX=minX=X[1];
        maxY=minY=Y[1];

        for(int i=2;i<=N;++i)
        {
            cin>>X[i]>>Y[i];

            if(maxX<X[i])
                maxX=X[i];

            if(maxY<Y[i])
                maxY=Y[i];

            if(minX>X[i])
                minX=X[i];

            if(minY>Y[i])
                minY=Y[i];
        }

        cout<<"Case #"<<cas<<": ";
        slove(maxX+maxY-minX-minY);
    }

    return 0;
}