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 (-10
9 ≤ x
i, y
i ≤ 10
9), 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
首先二分最大距离，对于每个距离采用舞蹈链算法判断其是否满足要求
代码如下
1 #include<bits/stdc++.h>
2 #define FOR(i,p,X) for(int i=X[p];i!=p;i=X[i])
3 #define For(i,a,b) for(int i=(a),i_end=(b);i<=i_end;++i)
4 using namespace std;
5 const int N=70;
6 int n,m,ans;
7 vector<int>G[N];
8 struct DLX{
9 int L[N*N],R[N*N],U[N*N],D[N*N];
10 int C[N*N],H[N],cnt[N],vis[N],id;
11 void init(){
12 For(i,0,n){
13 cnt[i]=0;U[i]=D[i]=i;
14 L[i+1]=i;R[i]=i+1;
15 }
16 R[n]=0;id=n+1;
17 memset(H,-1,sizeof(H));
18 }
19 void Link(int r,int c){
20 cnt[c]++;C[id]=c;
21 U[id]=U[c];D[U[c]]=id;
22 D[id]=c;U[c]=id;
23 if(!~H[r]) H[r]=L[id]=R[id]=id;
24 else{
25 L[id]=L[H[r]];R[L[H[r]]]=id;
26 R[id]=H[r];L[H[r]]=id;
27 }
28 id++;
29 }
30 void Remove(int sz){
31 FOR(j,sz,D)L[R[j]]=L[j],R[L[j]]=R[j];
32 }
33 void Resume(int sz){
34 FOR(j,sz,D)L[R[j]]=R[L[j]]=j;
35 }
36 int h(){
37 int res=0;
38 memset(vis,0,sizeof(vis));
39 FOR(i,0,R){
40 if(vis[i])continue;
41 ++res;
42 FOR(j,i,D)FOR(k,j,R)
43 vis[C[k]]=1;
44 }
45 return res;
46 }
47 bool Dance(int k){
48 if(k+h()>m)return false;
49 int pos=R[0];
50 if(!pos)return k<=m;
51 FOR(i,0,R)if(cnt[pos]>cnt[i])pos=i;
52 FOR(i,pos,D){
53 Remove(i);
54 FOR(j,i,R)Remove(j);
55 if(Dance(k+1))return true;
56 FOR(j,i,R)Resume(j);
57 Resume(i);
58 }
59 return false;
60 }
61 }dlx;
62 struct Point{
63 int x,y;
64 void input(){
65 scanf("%d%d",&x,&y);
66 }
67 }city[N];
68 long long dis(Point a,Point b){
69 return (long long)abs(a.x-b.x)+(long long)abs(a.y-b.y);
70 }
71 int main(){
72 int T;
73 scanf("%d",&T);
74 For(iCase,1,T){
75 scanf("%d%d",&n,&m);
76 For(i,1,n)city[i].input();
77 long long l=0,r=100000000000LL;
78 long long ans=0;
79 while(l<=r){
80 long long mid=l+r>>1;
81 dlx.init();
82 For(i,1,n)For(j,1,n)
83 if(dis(city[i],city[j])<=mid)
84 dlx.Link(i,j);
85 if(dlx.Dance(0))r=(ans=mid)-1;
86 else l=mid+1;
87 }
88 printf("Case #%d: %I64d\n",iCase,ans);
89 }
90 return 0;
91 }