HDU2295-Radar

Radar

                                                                    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                           Total Submission(s): 4150    Accepted Submission(s): 1592

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

13 3 23 43 15 41 12 23 3

 

Sample Output

2.236068

 

Source

The 4th Baidu Cup final

 

Recommend

lcy

 

题意：有n个城市，m个雷达,k个操作员，每个操作员只能操作一个雷达。每个雷达的覆盖范围是一个以雷达坐标为中心的圆，所有雷达的覆盖半径是相同的。现在给出这n个城市,m个雷达的坐标，问雷达覆盖半径最小是多少，让所有城市都可以被雷达覆盖到

解题思路：二分+舞蹈链重复覆盖

#include <iostream>    #include <cstdio>    #include <cstring>    #include <string>    #include <algorithm>    #include <cctype>    #include <map>    #include <cmath>    #include <set>    #include <stack>    #include <queue>    #include <vector>    #include <bitset>    #include <functional>    using namespace std;#define LL long long    const int INF = 0x3f3f3f3f;const int maxn = 300005;int n, m, K, x[1009], y[1009];struct DLX{int L[maxn], R[maxn], U[maxn], D[maxn];int row[maxn], col[maxn], sum[maxn], ans[maxn];int n, m, num, cnt;int vis[maxn], flag[maxn];void add(int k, int l, int r, int u, int d, int x, int y){L[k] = l;   R[k] = r;   U[k] = u;D[k] = d;   row[k] = x;  col[k] = y;}void reset(int n, int m){this->n = n;   this->m = m;for (int i = 0; i <= m; i++){add(i, i - 1, i + 1, i, i, 0, i);sum[i] = 0;}L[0] = m, R[m] = 0, cnt = m + 1;}void insert(int x, int y){int temp = cnt - 1;if (row[temp] != x){add(cnt, cnt, cnt, U[y], y, x, y);U[D[cnt]] = cnt; D[U[cnt]] = cnt;}else{add(cnt, temp, R[temp], U[y], y, x, y);R[L[cnt]] = cnt; L[R[cnt]] = cnt;U[D[cnt]] = cnt; D[U[cnt]] = cnt;}sum[y]++, cnt++;}void Remove(int k){for (int i = D[k]; i != k; i = D[i]){L[R[i]] = L[i];R[L[i]] = R[i];}}void Resume(int k){for (int i = U[k]; i != k; i = U[i]) L[R[i]] = R[L[i]] = i;}int A(){int dis = 0;for (int i = R[0]; i != 0; i = R[i]) vis[i] = 0;for (int i = R[0]; i != 0; i = R[i])if (!vis[i]){dis++, vis[i] = 1;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 dis;}void Dfs(int k){if (!R[0]) num = min(num, k);else if (k + A() < num){int now = R[0];for (int i = R[0]; i != 0; i = R[i])if (sum[now] > sum[i]) now = i;for (int i = D[now]; i != now; i = D[i]){Remove(i);for (int j = R[i]; j != i; j = R[j]) Remove(j);Dfs(k + 1);for (int j = L[i]; j != i; j = L[j]) Resume(j);Resume(i);}}}void mul(){memset(flag, 0, sizeof flag);num = 0x7FFFFFFF;}}dlx;int check(double k){dlx.reset(m, n);for (int i = 1; i <= m; i++)for (int j = 1; j <= n; j++){double temp = (x[n + i] - x[j])*(x[n + i] - x[j]) + (y[n + i] - y[j])*(y[n + i] - y[j]);if (k - sqrt(temp) > 1e-8) dlx.insert(i, j);}dlx.mul();dlx.Dfs(0);if (dlx.num <= K) return 1;return 0;}int main(){int t;scanf("%d", &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 = 0, r = 1500;while (r - l > 1e-8){double mid = (l + r) / 2;if (check(mid)) r = mid;else l = mid;}printf("%.6f\n", r);}return 0;}