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HDU 5046 Airport（DLX可重复覆盖）

2015-01-25 07:46:43 · 作者: · 浏览: 4
标签: HDU 5046 Airport DLX 重复 覆盖
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 (-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
题意：要在n个城市里建造不超过k个机场，问机场城市之间最大距离最小为多少。 DLX做法：跟

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。只不过我的方法快点。

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               using namespace std; #define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i ) #define REP( i , n ) for ( int i = 0 ; i < n ; ++ i ) #define CLEAR( a , x ) memset ( a , x , sizeof a ) typedef long long LL; typedef pair
              
               pil; const int maxn = 60+5; const int maxnnode=maxn*maxn; const int mod = 1000000007; int K; struct DLX{ int n,m,size; int U[maxnnode],D[maxnnode],L[maxnnode],R[maxnnode],Row[maxnnode],Col[maxnnode]; int H[maxn],S[maxn];//H[i]位置，S[i]个数 int ansd; void init(int a,int b) { n=a; m=b; REPF(i,0,m) { S[i]=0; U[i]=D[i]=i; L[i]=i-1; R[i]=i+1; } R[m]=0; L[0]=m; size=m; REPF(i,1,n) 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[maxn]; 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; } bool Dance(int d) { if(d + f() >K) return false; if(R[0] == 0) return d<=K; 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)) return true; for(int j = L[i];j != i;j = L[j])resume(j); resume(i); } return false; } }; struct point{ LL x,y; }X[maxn]; LL d[maxn][maxn]; LL dd[maxn*maxn]; DLX L; int t,n,cnt; int cas=1; LL dis(point a,point b) { return abs(a.x-b.x)+abs(a.y-b.y); } void solve() { int l=0,r=cnt-1; // int ans; while(l<=r) { int mid=(l+r)>>1; L.init(n,n); REPF(i,1,n) { REPF(j,1,n) if(d[i][j]<=dd[mid]) L.link(i,j); } if(L.Dance(0)) r=mid-1; else l=mid+1; } printf("Case #%d: %I64d\n",cas++,dd[l]); } int main() { LL x,y; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&K); REPF(i,1,n) { scanf("%I64d%I64d",&x,&y); X[i].x=x;X[i].y=y; } cnt=0; REPF(i,1,n) { REPF(j,1,n) { d[i][j]=dis(X[i],X[j]); dd[cnt++]=d[i][j]; } } sort(dd,dd+cnt); solve(); } return 0; }