HDU5046-Airport

Airport

                                                                      Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                              Total Submission(s): 2247    Accepted Submission(s): 711

Problem Description

The country of jiuye composed by N cites. Each city can be viewed as a point in a two- dimensional plane with integer coordinates (x,y). The distance between city i and city j is defined by d_ij = |x_i - x_j| + |y_i - y_j|. jiuye want to setup airport in K cities among N cities. So he need your help to choose these K cities, to minimize the maximum distance to the nearest airport of each city. That is , if we define d_i(1 ≤ i ≤ N ) as the distance from city i to the nearest city with airport. Your aim is to minimize the value max{d_i|1 ≤ i ≤ N }. You just output the minimum.

 

Input

The first line of the input is T (1 ≤ T ≤ 100), which stands for the number of test cases you need to solve.

The first line of each case contains two integers N ,K (1 ≤ N ≤ 60,1 ≤ K ≤ N ),as mentioned above.

The next N lines, each lines contains two integer x_i and y_i (-10⁹ ≤ x_i, y_i ≤ 10⁹), denote the coordinates of city i.

 

Output

For each test case, print a line “Case #t: ”(without quotes, t means the index of the test case) at the beginning. Then a single integer means the minimum.

 

Sample Input

23 20 04 05 14 20 31 03 08 9

 

Sample Output

Case #1: 2Case #2: 4

 

Source

2014 ACM/ICPC Asia Regional Shanghai Online

 

题意：有n个城市，要再其中k个城市中建机场，问每个城市到机场最近距离的最大值的最小值是多少

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 LL INF = 0x3f3f3f3f3f3f3f3f;const int maxn = 300005;int n, K;LL 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];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;}bool Dfs(int k){if (!R[0]) { num = min(num, k); return true; }else if (k + A() > K) { num = 0x7FFFFFFF; return false; }else{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);if (Dfs(k + 1)) return true;for (int j = L[i]; j != i; j = L[j]) Resume(j);Resume(i);}}return false;}void mul(){num = 0x7FFFFFFF;}}dlx;int check(LL k){dlx.reset(n, n);for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++){LL temp = abs(x[i] - x[j]) + abs(y[i] - y[j]);if (k >= temp) dlx.insert(i, j);}dlx.mul();dlx.Dfs(0);if (dlx.num <= K) return 1;return 0;}int main(){int t, cas = 0;scanf("%d", &t);while (t--){scanf("%d%d", &n, &K);for (int i = 1; i <= n; i++) scanf("%lld%lld", &x[i], &y[i]);LL l = 0, r = 0 ,ans;for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)r = max(r, abs(x[i] - x[j]) + abs(y[i] - y[j]));while (r >= l){LL mid = (l + r) / 2;if (check(mid)) r = mid-1, ans = mid;else l = mid + 1;}printf("Case #%d: %lld\n", ++cas, ans);}return 0;}