Radar HDU - 2295 DLX,重复覆盖+剪枝

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.

InputThe 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
OutputFor each test case, output the radius on a single line, rounded to six fractional digits.Sample Input

Sample Output

2.236068

翻译：爪哇王国的N个城市需要被雷达覆盖，处于战争状态。因为王国有M个雷达站，但只有K个操作员，所以我们最多可以操作K个雷达。所有雷达都有相同的圆形覆盖半径R。我们的目标是尽量减少R，同时覆盖整个城市不超过K雷达。

输入

输入由几个测试用例组成。输入的第一行由一个整数t组成，表示测试用例的数量。每个测试用例的第一行由3个整数组成：n、m、k，代表城市数量、雷达站数量和操作员数量。以下n条线中的每一条都由城市坐标组成。

最后的M线都由一个雷达站的坐标组成。

所有坐标用一个空格分隔。

产量

对于每个测试用例，在一行上输出半径，四舍五入为六位小数。

思路；雷达的半径用二分法求出，用搜索来检验当前雷达的半径下能否覆盖全部城市，用dlx模板优化搜索，dlx模板需要一些变化。

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 1000000007
#define eps 1e-6
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;

const  int maxm = 3010;
const double ep =1e-8;
int k;
struct{
    int n,m;
    int left[maxm],right[maxm],up[maxm],down[maxm],col[maxm],row[maxm];
    int head[55],ans[55];
    int ansed,size;
    void init(int _n,int _m)
    {
        n=_n;
        m=_m;
        for(int i=0;i<=m;i++)
        {
            ans[i]=0;
            up[i]=i;
            down[i]=i;
            left[i]=i-1;
            right[i]=i+1;
            col[i]=i;
            row[i]=0;
        }
        left[0]=m;
        right[m]=0;
        size=m;
        memset(head,-1,sizeof(head));
        return ;
    }
    void insert(int r,int c)
    {
        size++;
        ans[c]++;
        down[size]=down[c];
        up[size]=c;
        up[down[c]]=size;
        down[c]=size;
        row[size]=r;
        col[size]=c;
        if(head[r]<0)
        {
            head[r]=size;
            left[size]=right[size]=size;
        }
        else
        {
            left[size]=head[r];
            right[size]=right[head[r]];
            left[right[head[r]]]=size;
            right[head[r]]=size;
        }
        return ;
    }
    void delet(int c)//只删除一列
    {
        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 ve[maxm];
    int f()//还需要多少个操纵人员
    {
        int js=0;
        for(int i=right[0];i!=0;i=right[i])
        {
            ve[i]=true;
        }
        for(int c=right[0];c!=0;c=right[c])
        {
            if(ve[c])
            {
                js++;
                ve[c]=false;
                for(int i=down[c];i!=c;i=down[i])
                {
                    for(int j=right[i];j!=i;j=right[j])
                    {
                        ve[col[j]]=false;
                    }
                }
            }
        }
        return js;
    }
    bool dancing(int dep)
    {
        if(dep+f()>k)//剪枝
        {
            return false;
        }
        if(right[0]==0)
        {
            return dep<=k;
        }
        int c=right[0];
        for(int i=right[0];i!=0;i=right[i])
        {
            if(ans[i]<ans[c])
            {
                c=i;
            }
        }
        for(int i=down[c];i!=c;i=down[i])
        {
            delet(i);
            for(int j=right[i];j!=i;j=right[j])
            {    
                delet(j);
            }
            if(dancing(dep+1))
            {
                return true;
            }
            for(int j=left[i];j!=i;j=left[j])
            {
                reback(j);
            }
            reback(i);
        }
        return false;
    }
}dlx;
struct node
{
    int x,y;
    void inp()
    {
        scanf("%d %d",&x,&y);
    }
};
node city[55],station[55];
double dis(node a,node b)//求距离
{
    return sqrt((double)(a.x-b.x)*(a.x-b.x)+(double)(a.y-b.y)*(a.y-b.y));
}
int main()
{
    int n,m,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d %d",&n,&m,&k);
        for(int i=0;i<n;i++)
        {
            city[i].inp();
        }
        for(int i=0;i<m;i++)
        {
            station[i].inp();
        }
        double l=0,r=1e8;
        while(r-l>=ep)//二分
        {
            double mid=(l+r)/2;
            dlx.init(m,n);
            for(int i=0;i<m;i++)
            {
                for(int j=0;j<n;j++)
                {
                    if(dis(city[j],station[i])<mid-ep)//如果雷达能覆盖城市，则插入
                    {
                        dlx.insert(i+1,j+1);
                    }
                }
            }
            if(dlx.dancing(0))
            {
                r=mid-ep;
            }
            else
            {
                l=mid+ep;
            }
        }
        printf("%.6lf\n",l);
    }
}

Radar HDU - 2295 DLX,重复覆盖+剪枝

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