HDU 2295 Radar

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
   
   
   
   

二分+dlx重复覆盖验证
#include<cstdio>
#include<vector>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll maxn = 1005;
int n, m, x[maxn], y[maxn], T, k;

inline void read(int &ret)
{
    char c;
    do {
        c = getchar();
    } while (c < '0' || c > '9');
    ret = c - '0';
    while ((c = getchar()) >= '0' && c <= '9')
        ret = ret * 10 + (c - '0');
}

struct DLX
{
    #define maxn 10005
    #define F(i,A,s) for (int i=A[s];i!=s;i=A[i])
    int L[maxn], R[maxn], U[maxn], D[maxn];
    int row[maxn], col[maxn], ans[maxn], cnt[maxn];
    int n, m, num, sz;

    bool Flag;
    void add(int now, int l, int r, int u, int d, int x, int y)
    {
        L[now] = l;    R[now] = r;    U[now] = u;
        D[now] = d;   row[now] = x;  col[now] = y;
    }
    void reset(int n, int m)
    {
        Flag = false;
        this->n = n;    this->m = m;
        for (int i = 0; i <= m; i++)
        {
            add(i, i - 1, i + 1, i, i, 0, i);
            cnt[i] = 0;
        }
        L[0] = m;     R[m] = 0;     sz = m + 1;
    }
    void insert(int x, int y)
    {
        int ft = sz - 1;
        if (row[ft] != x)
        {
            add(sz, sz, sz, U[y], y, x, y);
            U[D[sz]] = sz; D[U[sz]] = sz;
        }
        else
        {
            add(sz, ft, R[ft], U[y], y, x, y);
            R[L[sz]] = sz; L[R[sz]] = sz;
            U[D[sz]] = sz; D[U[sz]] = sz;
        }
        ++cnt[y];    ++sz;
    }

    //精确覆盖
    void remove(int now)
    {
        R[L[now]] = R[now];
        L[R[now]] = L[now];
        F(i,D,now) F(j,R,i)
        {
            D[U[j]] = D[j];
            U[D[j]] = U[j];
            --cnt[col[j]];
        }
    }
    void resume(int now)
    {
        F(i,U,now)    F(j,L,i)
        {
            D[U[j]] = j;
            U[D[j]] = j;
            ++cnt[col[j]];
        }
        R[L[now]] = now;
        L[R[now]] = now;
    }
    int dfs(int x)
    {
        if (!R[0]) return 1;
        int now = R[0];
        F(i,R,0) if (cnt[now]>cnt[i]) now = i;
        remove(now);
        F(i,D,now)
        {
            ans[x] = row[i];
            F(j,R,i) remove(col[j]);
            if (dfs(x + 1)) return 1;
            F(j,L,i) resume(col[j]);
        }
        resume(now);
        return 0;
    }

    //精确覆盖
    //重复覆盖
    void Remove(int now)
    {
        F(i,D,now) 
        {
            L[R[i]]=L[i];
            R[L[i]]=R[i];
        }
    }
    void Resume(int now)
    {
        F(i,U,now) L[R[i]]=R[L[i]]=i;
    }
    int vis[maxn];
    int flag[maxn];
    int A()
    {
        int dis=0;
        F(i,R,0) vis[i]=0;
        F(i,R,0) if (!vis[i])
        {
            dis++;    vis[i]=1;
            F(j,D,i) F(k,R,j) vis[col[k]]=1;
        }
        return dis;
    }
    void Dfs(int x)
    {
        if (!R[0]) num=min(num,x);
        else if (x+A()<num)
        {
            int now=R[0];
            F(i,R,0) if (cnt[now]>cnt[i]) now = i;
            F(i,D,now)
            {
                Remove(i);F(j,R,i) Remove(j);
                Dfs(x+1);
                F(j,L,i) Resume(j);Resume(i);
            }
        }
    }
    void mul()
    {
        num=0x7FFFFFFF;
    }
    //重复覆盖
}dlx;

void reset(double xx)
{
    dlx.reset(m,n);
    for (int i=1;i<=m;++i)
    {
        for (int j=1;j<=n;++j)
        {
            double u=(x[n+i]-x[j])*(x[n+i]-x[j])+(y[n+i]-y[j])*(y[n+i]-y[j]);
            if (xx-sqrt(u)>1e-8) dlx.insert(i,j);
        }
    }
}

bool check()
{
    dlx.mul();
    dlx.Dfs(0);
    if (dlx.num<=k) return true;
    else return false;
}

int main()
{
    read(T);
    while (T--)
    {
        scanf("%d%d%d",&n,&m,&k);
        for (int i=1;i<=n+m;++i) scanf("%d%d",&x[i],&y[i]);
        double l,r,mid;
        for (l=0,r=1500;r-l>1e-8;)
        {
            mid=(l+r)/2;
            reset(mid);
            if (check()) r=mid; else l=mid;
        }
        printf("%.6lf\n",r);
    }
    return 0;
}