Airport（未解决。。。）

Airport

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1534    Accepted Submission(s): 486

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

2 3 2 0 0 4 0 5 1 4 2 0 3 1 0 3 0 8 9

 

 

Sample Output

Case #1: 2 Case #2: 4

 

题意：就是给你一系列点，让你挑k个点建造飞机场，然后求剩下点到最近的飞机场的最小的最大距离；距离定义：d_ij = |x_i - x_j| + |y_i - y_j|

我的思路是分别找出每个点到剩下点的距离，然后sort排序，找出每个城市第k大值的最小值；

大婶们说这是DLX可重复覆盖：

下面是大神的说法：

HDU 2295 Radar（DLX可重复覆盖）差不多，我们的做法就是

保存n个城市之间的距离，sort一下，二分结果，对满足条件的DLX求覆盖程度，

求出最大距离最小值。此题二分0~INF也可解决。

http://www.2cto.com/kf/201412/364538.html

先把我的wa代码扔着，抽空补上。。。

wa代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
const int MAXN=1e5+10;
int a[65][65];
int cmp(int a,int b){
	return a>b;
}
struct Node{
	int x,y;
};
Node dt[65];
int getl(Node a,Node b){
	return abs(a.x-b.x)+abs(a.y-b.y);
}
int main(){
	int T,N,K,kase=0;
	scanf("%d",&T);
	while(T--){
		scanf("%d%d",&N,&K);
		for(int i=0;i<N;i++){
			scanf("%d%d",&dt[i].x,&dt[i].y);
		}
		for(int i=0;i<N;i++)for(int j=0;j<N;j++){
			a[i][j]=getl(dt[i],dt[j]);
		}
		for(int i=0;i<N;i++)sort(a[i],a[i]+N,cmp);
	/*	for(int i=0;i<N;i++){
			for(int j=0;j<N;j++)printf("%d ",a[i][j]);
			puts("");
		}*/
		int mx=0x3f3f3f3f;
		for(int i=0;i<N;i++)mx=min(mx,a[i][K-1]);
		printf("Case #%d: %d\n",++kase,mx);
	}
	return 0;
}