DLX+hdu2295_n cities of the java kingdom need to be covered…-优快云博客

本文链接：https://blog.youkuaiyun.com/u010660276/article/details/39647697

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Radar

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2500 Accepted Submission(s): 974

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

Sample Output

      
       2.236068

从m个雷达站中选出k个覆盖n个城市

DLX，原来的模板不好用，老是出bug，还是用kuangbin大神的吧

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
const int maxn=60;
const int maxnode=60*60;
const int maxr=60;
const double eps=1e-8;
int N,M,K;
struct node
{
    double x,y;
}p[maxn],p1[maxn];

struct DLX
{
    int n,m,size;
    int U[maxnode],D[maxnode],R[maxnode],L[maxnode],Row[maxnode],Col[maxnode];
    int H[maxn],S[maxn];
    int ands,ans[maxn];
    void init(int _n,int _m)
    {
        n = _n;
        m = _m;
        for(int i = 0;i <= m;i++)
        {
            S[i] = 0;
            U[i] = D[i] = i;
            L[i] = i-1;
            R[i] = i+1;
        }
        R[m] = 0; L[0] = m;
        size = m;
        for(int i = 1;i <= n;i++)
            H[i] = -1;
    }
    void Link(int r,int c)
    {
        ++S[Col[++size]=c];
        Row[size] = r;
        D[size] = D[c];
        U[D[c]] = size;
        U[size] = c;
        D[c] = size;
        if(H[r] < 0)H[r] = L[size] = R[size] = size;
        else
        {
            R[size] = R[H[r]];
            L[R[H[r]]] = size;
            L[size] = H[r];
            R[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 v[maxnode];
    int f()
    {
        int ret = 0;
        for(int c = R[0];c != 0;c = R[c])v[c] = true;
        for(int c = R[0];c != 0;c = R[c])
            if(v[c])
            {
                ret++;
                v[c] = false;
                for(int i = D[c];i != c;i = D[i])
                    for(int j = R[i];j != i;j = R[j])
                        v[Col[j]] = false;
            }
        return ret;

    }
    bool Dance(int d)
    {
        if(d + f() > K)return false;
        if(R[0] == 0)return d <= K;
        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 true;
            for(int j = L[i];j != i;j = L[j])resume(j);
            resume(i);
        }
        return false;
    }
}dlx;

double D(int i,int j)
{
    double x=p1[i].x-p[j].x;
    double y=p1[i].y-p[j].y;
    return sqrt(x*x+y*y);
}
bool can(double R)
{
    dlx.init(M,N);
    for(int i=1;i<=M;i++)
    {
        for(int j=1;j<=N;j++)
            if(D(i,j)<R-eps)dlx.Link(i,j);
    }
    return dlx.Dance(0);
}
void solve()
{
    double l=0,r=1e8,mid;
    while(r-l>eps)
    {
        mid=(l+r)/2;
        if(can(mid))r=mid-eps;
        else l=mid+eps;
    }
    printf("%.6lf\n",l);
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&N,&M,&K);
        for(int i=1;i<=N;i++)scanf("%lf%lf",&p[i].x,&p[i].y);
        for(int i=1;i<=M;i++)scanf("%lf%lf",&p1[i].x,&p1[i].y);
        solve();
    }
    return 0;
}