思路
- 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
- 二分枚举答案
- 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)
- 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
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()
{
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()
{
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;
}