B - Radar HDU - 2295（二分+重复覆盖）

题目传送门

思路

• 题意

1. n个城市，m个雷达站，从m个雷达站中选择k个使n个城市到打雷达站的最大距离最小，求出这个最大最小值

• 思路

2. 二分枚举答案
3. check（int md）函数中用 “DLX重复覆盖”来暴力枚举选择k个城市看在二分枚举的距离为md 的情况下能否覆盖所有城市（可以重复覆盖）
4. 对与构造DLX重复覆盖所需要的 数据矩阵 行表示 m个雷达章，列表示n个将被覆盖的城市。行和列的交点元素为1，如果在md距离下行所代表的雷达站能够覆盖列所代表的城市；否则交点元素为0

详细分析传送门

代码

#include<iostream> 
#include<algorithm>
#include<cstring>
#include<queue>
#include<map>
#include<string>
#include<cstdio>
#include<cmath>
#include<stack>
void fre() { system("clear"), freopen("A.txt", "r", stdin); freopen("Ans.txt","w",stdout); } 
void Fre() { system("clear"), freopen("A.txt", "r", stdin);}
#define ios ios::sync_with_stdio(false)
#define Pi acos(-1)
#define pb push_back
#define fi first
#define se second
#define ll long long
#define ull unsigned long long 
#define db double
#define Pir pair<int, int>
#define PIR pair<Pir, Pir>
#define m_p make_pair
#define INF 0x3f3f3f3f
#define esp 1e-7
#define mod (ll)(1e9 + 7)
#define for_(i, s, e) for(int i = (ll)(s); i <= (ll)(e); i ++)
#define rep_(i, e, s) for(int i = (ll)(e); i >= (ll)(s); i --)
#define sc scanf
#define sd(a) scanf("%d", &a)
#define ss(a) scanf("%s", a)
using namespace std;

const int mxn = 55;
const int mxnode = 3000;
int n, m, k;
db dis[mxn][mxn];
pair<db, db> city[mxn], rada[mxn];

struct DLX
{
    int U[mxnode], D[mxnode], R[mxnode], L[mxnode];
    int Col[mxnode], Row[mxnode];
    int H[mxnode], S[mxnode];
    int ans, ansd[mxn];
    int size, n;
    int vis[mxn];

    void init(int _n)
    {
        n = _n;
        for_(i, 0, n)
        {
            L[i] = i - 1;
            R[i] = i + 1;
            U[i] = D[i] = i;
        }
        R[n] = 0, L[0] = n, size = n;
        memset(H, -1, sizeof(H));
        memset(S,  0, sizeof(S));
    }
    void link(int r, int c)
    {
        Col[++ size] = c;
        Row[size] = r;
        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] = size - 1;
            R[size] = H[r];
            R[L[size]] = 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]] = i, R[L[i]] = i;
    }
    int h()            //类似与A* 算法中的 h估值函数，表示从当前状态到结束状态至少还需要递归多少层；
    {
        memset(vis, 0, sizeof(vis));
        int sum = 0;
        for(int i = R[0]; i; i = R[i])
        {
            if(vis[i]) continue;
            vis[i] = 1;
            sum ++;
            for(int j = D[i]; j != i; j = D[j])
                for(int k = R[j]; k != j; k = R[k])
                    vis[Col[k]] = 1;
        }
        return sum;
    }
    bool dance(int d)
    {
        if(d + h() > k) return false;
        if(R[0] == 0) return d <= k;

        int c = R[0];
        for(int i = R[0]; i; i = R[i])
            if(S[c] > S[i])
                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;

db Dis(int i, int j)
{
    db x = abs(city[j].fi - rada[i].fi);
    db y = abs(city[j].se - rada[i].se);
    return sqrt(x * x + y * y);
}


bool check(db md)
{
    dlx.init(n);
    for_(i, 1, m)
        for_(j, 1, n)
        if(dis[i][j] <= md)
            dlx.link(i, j);
    return dlx.dance(0);
}


int main()
{
    /* fre(); */
    int T;
    sd(T);
    while(T --)
    {
        sc("%d %d %d", &n, &m, &k);    
        for_(i, 1, n)
            sc("%lf %lf", &city[i].fi, &city[i].se);
        for_(i, 1, m)
        {
            sc("%lf %lf", &rada[i].fi, &rada[i].se);
            for_(j, 1, n)
                dis[i][j] = Dis(i, j);
        }
        db l = 0, r = 2000;
        while(r - l >= esp)
        {
            db md = (l + r) / 2;
            if(check(md))
            {
                r = md;
            }
            else
            {
                l = md;
            }
        }
        printf("%.6f\n", l);
    }

    return 0;
}