hdu 5046 Airport (重复覆盖)

Airport

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1481    Accepted Submission(s): 469

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

二分 + 重复覆盖

/*======================================================# Author: whai# Last modified: 2015-10-09 11:30# Filename: hdu5046.cpp======================================================*/#include <iostream>#include <cstdio>#include <vector>#include <algorithm>#include <cstring>#include <string>#include <cmath>#include <set>#include <map>using namespace std;#define LL __int64#define PB push_back#define P pair<int, int>#define X first#define Y secondconst int MAXM = 65;const int MAXN = 65;const int N = MAXN * MAXM;const int INF = 0x3f3f3f3f;struct DLX {    int n, m, size;    int U[N], D[N], R[N], L[N], row[N], col[N];    int H[MAXN], S[MAXM];    int ans[MAXN], ans_cnt;    void init(int _n, int _m) {        ans_cnt = INF;        n = _n;        m = _m;        for (int i = 0; i <= m; i++) {            S[i] = 0;            U[i] = D[i] = i;            L[i] = i - 1;            R[i] = i + 1;        }        R[m] = 0; L[0] = m;        size = m;        for (int i = 1; i <= n; i++)H[i] = -1;    }    void link(int r, int c) {        ++S[col[++size] = c];        row[size] = r;        D[size] = D[c];        U[D[c]] = size;        U[size] = c;        D[c] = size;        if (H[r] < 0)H[r] = L[size] = R[size] = size;        else {            R[size] = R[H[r]];            L[R[H[r]]] = size;            L[size] = H[r];            R[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]] = R[L[i]] = i;    }    bool v[MAXM];    int f() {        int ret = 0;        for (int c = R[0]; c != 0; c = R[c])v[c] = true;        for (int c = R[0]; c != 0; c = R[c])            if (v[c]) {                ret++;                v[c] = false;                for (int i = D[c]; i != c; i = D[i])                    for (int j = R[i]; j != i; j = R[j])                        v[col[j]] = false;            }        return ret;    }//注意这些可以加入剪枝优化的    /*void dance(int d) {        if (d + f() >= ans_cnt) return;        if (R[0] == 0) {            if (d < ans_cnt) ans_cnt = d;            return ;        }        int c = R[0];        for (int i = R[0]; i != 0; i = R[i])            if (S[i] < S[c])                c = i;        for (int i = D[c]; i != c; i = D[i]) {            remove(i);            //ans[d] = row[i];            for (int j = R[i]; j != i; j = R[j]) remove(j);            dance(d + 1);            for (int j = L[i]; j != i; j = L[j]) resume(j);            resume(i);        }    }*/bool dance(int d, int limit) {if (d + f() > limit) return false;        if (R[0] == 0) return d <= limit;        int c = R[0];        for (int i = R[0]; i != 0; i = R[i])            if (S[i] < S[c])                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, limit)) return true;            for (int j = L[i]; j != i; j = L[j]) resume(j);            resume(i);        }return false;    }} dlx;P a[N];LL _abs(int x) {    if(x < 0) return -x;    return x;}LL dis(P a, P b) {    return _abs(a.X - b.X) + _abs(a.Y - b.Y);}bool ok(LL d, int n, int k) {    dlx.init(n, n);    for(int i = 0; i < n; ++i) {        for(int j = 0; j < n; ++j) {            if(dis(a[i], a[j]) <= d) {                dlx.link(i + 1, j + 1);            }        }    }    return dlx.dance(0, k);}void gao(int n, int k) {    LL L = 0, R = 4 * 1e9, ans = 4 * 1e9;    while(L < R) {        LL mid = (L + R) >> 1;        if(!ok(mid, n, k)) {            L = mid + 1;        } else {            ans = min(ans, mid);            R = mid;        }    }    if(ok(L, n, k)) ans = min(ans, L);    if(ok(R, n, k)) ans = min(ans, R);    printf("%I64d\n", ans);}int main() {    int T;    int cas = 0;    scanf("%d", &T);    while(T--) {        int n, k;        scanf("%d%d", &n, &k);        for(int i = 0; i < n; ++i) {            scanf("%d%d", &a[i].X, &a[i].Y);        }        printf("Case #%d: ", ++cas);        gao(n, k);    }    return 0;}