hdoj Airport 5046 （） 好题

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

23 20 04 05 14 20 31 03 08 9

Sample Output

Case #1: 2Case #2: 4
 
唉，写了一早上，都没写出来。。
先贴个错的代码，明天再看
#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>#define INF 0x3f3f3f3f#define ll long longusing namespace std;ll sum[110];ll a[110][110];struct zz{ll x;ll y;}q[1010];int main(){int t,T=1;int n,m,i,j,k,kk;scanf("%d",&t);while(t--){scanf("%d%d",&n,&k);for(i=0;i<n;i++)scanf("%lld%lld",&q[i].x,&q[i].y);//sort(q,q+n,cmp);memset(a,0,sizeof(a));memset(sum,0,sizeof(sum));for(i=0;i<n;i++){kk=0;for(j=0;j<n;j++){a[i][kk++]=abs(q[i].x-q[j].x)+abs(q[i].y-q[j].y);}}for(i=0;i<n;i++)sort(a[i],a[i]+kk);for(i=0;i<n;i++){for(j=0;j<kk;j++){printf("%d ",a[i][j]);}printf("\n");}int m=0;for(i=0;i<n;i++)sum[m++]=a[i][kk-k];sort(sum,sum+m);printf("Case #%d: %lld\n",T++,sum[0]);}return 0;}