Airport HDU

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.

InputThe first line of the input is T (1 ≤ T ≤ 100), which stands for the number of test cases you need to solve. 

The first line of each case contains two integers N ,K (1 ≤ N ≤ 60,1 ≤ K ≤ N ),as mentioned above. 

The next N lines, each lines contains two integer x _i and y _i (-10 ⁹ ≤ x _i, y _i ≤ 10 ⁹), denote the coordinates of city i.OutputFor each test case, print a line “Case #t: ”(without quotes, t means the index of the test case) at the beginning. Then a single integer means the minimum.Sample Input

2
3 2
0 0
4 0
5 1
4 2
0 3
1 0
3 0
8 9

Sample Output

Case #1: 2
Case #2: 4

题意：有n块小岛，现在要在其中的k块小岛上建立飞机场，给出小岛的坐标，其中小岛之间的距离d(i,j)=|xi-xj|+|yi-yj|;求建立k个飞机时，飞机的最小航距时多少；

思路：二分法判断航距，用DLX重复覆盖判断当前航距下能否满足k座，最后一步步缩小答案。注意岛之间的距离非常大，用long long 存放。

代码：

#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <iostream>
#define mod 998244353
#define eps 1e-6
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;

const int maxn=4010;
int k;
struct 
{
    int left[maxn],right[maxn],up[maxn],down[maxn];
    int head[65],col[maxn];
    int num[65],id;
    void init(int m)
    {
        for(int i=0;i<=m;i++)
        {
            left[i]=i-1;
            right[i]=i+1;
            up[i]=down[i]=i;
            col[i]=i;
        }
        id=m;
        left[0]=m;
        right[m]=0;
        memset(head,-1,sizeof(head));
        memset(num,0,sizeof(num));
    }
    void link(int x,int y)
    {
        id++;
        down[id]=down[y];
        up[down[y]]=id;
        up[id]=y;
        down[y]=id;
        num[y]++;
        col[id]=y;
        if(head[x]==-1)
        {
            head[x]=left[id]=right[id]=id;
        }
        else
        {
            right[id]=right[head[x]];
            left[right[head[x]]]=id;
            left[id]=head[x];
            right[head[x]]=id;
        }
    }
    void remove(int c)
    {
        for(int i=down[c];i!=c;i=down[i])
        {
            left[right[i]]=left[i];
            right[left[i]]=right[i];
        }
    }
    void reback(int c)
    {
        for(int i=up[c];i!=c;i=up[i])
        {
            left[right[i]]=right[left[i]]=i;
        }
    }
    bool bj[maxn];
    int A()
    {
        int ans=0;
        for(int c=right[0];c!=0;c=right[c])
        {
            bj[c]=true;
        }
        for(int c=right[0];c!=0;c=right[c])
        {
            if(bj[c])
            {
                ans++;
                bj[c]=false;
                for(int i=down[c];i!=c;i=down[i])
                {
                    for(int j=right[i];j!=i;j=right[j])
                    {
                        bj[col[j]]=false;
                    }
                }
            }
        }
        return ans;
    }
    bool danc(int step)
    {
        if(A()+step>k)//剪枝
        {
            return false;
        }
        if(right[0]==0)
        {
            return step<=k;
        }
        int c=right[0];
        for(int i=c;i!=0;i=right[i])
        {
            if(num[i]<num[c])
            {
                c=i;
            }
        }
        for(int i=down[c];i!=c;i=down[i])
        {
            remove(i);
            for(int j=right[i];j!=i;j=right[j])
            {
                remove(j);
            }
            if(danc(step+1))
            {
                return true;
            }
            for(int j=left[i];j!=i;j=left[j])
            {
                reback(j);
            }
            reback(i);
        }
        return false;
    }
}dlx;
struct node//小岛信息
{
    ll x,y;
}no[65];
ll dis(node a,node b)//计算距离
{
    ll dx=a.x-b.x;
    if(dx<0)
    {
        dx=-dx;
    }
    ll dy=a.y-b.y;
    if(dy<0)
    {
        dy=-dy;
    }
    return dx+dy;
}
int main()
{
    int t,ans=0;;
    scanf("%d",&t);
    while(t--)
    {
        ans++;
        int n;
        ll x[65],y[65];
        scanf("%d %d",&n,&k);
        for(int i=1;i<=n;i++)
        {
            scanf("%lld %lld",&no[i].x,&no[i].y);
        }
        ll le=0,rig=100000000000LL;
        while(rig-le>0)
        {
            dlx.init(n);
            ll mid=(rig+le)/2;
            for(int i=1;i<=n;i++)
            {
                for(int j=1;j<=n;j++)
                {
                    if(dis(no[i],no[j])<=mid)//岛之间的距离比航距小时
                    {
                        dlx.link(i,j);
                    }
                }
            }
            if(dlx.danc(0))//如果当前航距需要的最少飞机的数量比k小为true
            {
                rig=mid;
            }
            else
            {
                le=mid+1;
            }
        }
        printf("Case #%d: %lld
",ans,le);
    }
}