HDU 5046 Airport

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
二分+dlx重复覆盖
#include<cstdio>#include<vector>#include<cmath>#include<map>#include<cstring>#include<algorithm>using namespace std;typedef long long ll;const ll maxn = 105;int T, n, m, x, y, t, tot, tt = 0;ll X[maxn], Y[maxn], mp[maxn][maxn], l, r, mid;inline void read(int &ret){char c;do {c = getchar();} while (c < '0' || c > '9');ret = c - '0';while ((c = getchar()) >= '0' && c <= '9')ret = ret * 10 + (c - '0');}struct DLX{#define maxn 500005#define F(i,A,s) for (int i=A[s];i!=s;i=A[i])int L[maxn], R[maxn], U[maxn], D[maxn];int row[maxn], col[maxn], ans[maxn], cnt[maxn];int n, m, num, sz;void add(int now, int l, int r, int u, int d, int x, int y){L[now] = l;R[now] = r;U[now] = u;D[now] = d;   row[now] = x;  col[now] = y;}void reset(int n, int m){num = 0x7FFFFFFF;this->n = n;this->m = m;for (int i = 0; i <= m; i++){add(i, i - 1, i + 1, i, i, 0, i);cnt[i] = 0;}L[0] = m; R[m] = 0; sz = m + 1;}void insert(int x, int y){int ft = sz - 1;if (row[ft] != x){add(sz, sz, sz, U[y], y, x, y);U[D[sz]] = sz; D[U[sz]] = sz;}else{add(sz, ft, R[ft], U[y], y, x, y);R[L[sz]] = sz; L[R[sz]] = sz;U[D[sz]] = sz; D[U[sz]] = sz;}++cnt[y];++sz;}//精确覆盖void remove(int now){R[L[now]] = R[now];L[R[now]] = L[now];F(i, D, now) F(j, R, i){D[U[j]] = D[j];U[D[j]] = U[j];--cnt[col[j]];}}void resume(int now){F(i, U, now)F(j, L, i){D[U[j]] = j;U[D[j]] = j;++cnt[col[j]];}R[L[now]] = now;L[R[now]] = now;}bool dfs(int x){//if (x + A() >= num) return;if (!R[0]) { num = min(num, x); return true; }int now = R[0];F(i, R, 0) if (cnt[now]>cnt[i]) now = i;remove(now);F(i, D, now){ans[x] = row[i];F(j, R, i) remove(col[j]);if (dfs(x + 1)) return true;F(j, L, i) resume(col[j]);}resume(now);return false;}//精确覆盖//重复覆盖void Remove(int now){F(i, D, now){L[R[i]] = L[i];R[L[i]] = R[i];}}void Resume(int now){F(i, U, now) L[R[i]] = R[L[i]] = i;}int vis[maxn];int flag[maxn];int A(){int dis = 0;F(i, R, 0) vis[i] = 0;F(i, R, 0) if (!vis[i]){dis++;vis[i] = 1;F(j, D, i) F(k, R, j) vis[col[k]] = 1;}return dis;}bool Dfs(int x, int y){if (x + A() > y) return false;if (!R[0]) return x <= y;else {int now = R[0];F(i, R, 0) if (cnt[now]>cnt[i]) now = i;F(i, D, now){Remove(i); F(j, R, i) Remove(j);if (Dfs(x + 1, y)) return true;F(j, L, i) Resume(j); Resume(i);}}return false;}//重复覆盖}dlx;int main(){read(T);while (T--){scanf("%d%d", &n, &m);for (int i = 1; i <= n; i++) scanf("%lld%lld", &X[i], &Y[i]);l = r = 0;for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++){mp[i][j] = abs(X[i] - X[j]) + abs(Y[i] - Y[j]);r = max(mp[i][j], r);}while (l <= r){mid = (l + r) >> 1;dlx.reset(n, n);for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)if (mp[i][j] <= mid) dlx.insert(i, j);if (dlx.Dfs(0, m)) r = mid - 1; else l = mid + 1;}printf("Case #%d: %lld\n", ++tt, l);}return 0;}