HDU 2295 (DLX + binary)

The processing of vertices discretization is used, and then the processing can be done in two minutes...

Radar

time limit: 2000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/others) total submission (s): 1928 accepted submission (s ): 796

Problem descriptionn cities of the Java Kingdom need to be covered by Radars for being in a state of war. since the Kingdom has m radar stations but only K operators, we can at most operate K radars. all radars have the same circular coverage with a radius of R. our goal is to minimize R while covering the entire city with no more than K radars.

 

Inputthe input consists of several test cases. the first line of the input consists of an integer T, indicating the number of test cases. the first line of each test case consists of 3 integers: n, m, K, representing the number of cities, the number of radar stations and the number of operators. each of the following n lines consists of the coordinate of a city. each of the last M lines consists of the coordinate of a radar station.
All coordinates are separated by one space. Technical Specification
1. 1 ≤ T ≤ 20 2. 1 ≤ n, m ≤ 50 3. 1 ≤ k ≤ M 4. 0 ≤ x, y ≤ 1000

 

Outputfor each test case, output the radius on a single line, rounded to six fractional digits.

 

Sample input1
3 3 2
3 4
3 1
5 4
1 1
2 2
3 3

 

Sample output2.236068

 

Sourcethe 4th Baidu Cup Final

 

Recommendlcy

# Include <stdio. h> # Include <Algorithm> # Include <String . H> # Include <Math. h> # Include <Iostream> Using   Namespace  STD;  # Define N 10000 # Define INF 0x3fffffff Struct  Node {  Double  X, Y;} GN [  55 ], GM [ 55 ];  Int  N, m, K;  Int  U [N], d [N], R [N], L [N], num [N], H [N], Col [N], line [N];  Int  Head, ID, mi;  Int  Nn, mm;  Double  G [N];  Double  Cal (node T, node T1 ){  Return SQRT (T. x-t1.x) * (T. x-t1.x) + (T. y-t1.y) * (T. Y- T1.y ));} Void  Prepare (){  For ( Int I = 0 ; I <= mm; I ++ ) {Num [I] = 0  ; U [I] = I; d [I] = I; R [I] = I + 1  ; L [I + 1 ] = I;} R [mm] = 0  ; L [  0 ] = Mm; memset (H, - 1 , Sizeof  (H ));}  Void Link ( Int TN, Int  TM) {ID ++ ; Num [line [ID] = Tm] ++ ; Col [ID] = TN; U [d [Tm] =ID; d [ID] = D [Tm]; U [ID] = TM; d [Tm] = ID;  If (H [tn] < 0 ) H [tn] = R [ID] = L [ID] = ID;  Else  {L [R [H [tn] = ID; R [ID] = R [H [tn]; L [ID] = H [tn]; R [H [tn] = ID ;}} Int  H (){  Int Mark [ 66  ]; Memset (mark,  0 , Sizeof  (Mark ));  Int Sum = 0  ;  For ( Int I = R [head]; I! = Head; I = R [I]) {  If (MARK [I] =0  ) {Sum ++ ; Mark [I] = 1  ;  For ( Int J = d [I]; J! = I; j = D [J])  For ( Int K = R [J]; k! = J; k = R [k]) MARK [line [k] = 1  ;}} Return  SUM ;}  Void Remove ( Int  S ){  For ( Int I = d [s]; I! = S; I = D [I]) {R [L [I] = R [I]; L [R [I] = L [I] ;}}  Void Resume ( Int  S ){  For (Int I = U [s]; I! = S; I = U [I]) R [L [I] = L [R [I] = I ;}  Void DFS ( Int  S ){  If (S + H ()> = mi) Return  ;  If (R [head] = Head) {mi = S;  Return  ;} Int TMI = INF, Tu;  For ( Int I = R [head]; I! = Head; I = R [I]) {  If (Num [I] < TMI) {TMI = Num [I]; TU = I ;}}  For ( Int I = d [Tu]; I! = Tu; I = D [I]) {remove (I );  For ( Int J = R [I]; J! = I; j = R [J]) Remove (j); DFS (S + 1  );  For ( Int J = L [I]; J! = I; j = L [J]) Resume (j); resume (I );}}  Int DLX ( Double  Key) {head = 0  ; Mm = N; NN =1  ; Prepare (); Id = Mm;  For ( Int I = 1 ; I <= m; I ++ ){  Int Flag = 0  ;  For ( Int J = 1 ; J <= N; j ++ ){ If (CAL (GN [J], GM [I]) <= Key) {flag = 1  ; Link (I, j );}}  If (Flag = 1 ) Nn ++ ;} Mi = INF; DFS (  0  );  If (MI <= K)  Return   1 ;  Else   Return   0  ;}  Int  Main (){  Int  T; scanf (  "  % D  " ,& T );  While (T -- ) {Scanf (  " % D  " , & N, & M ,& K );  For ( Int I = 1 ; I <= N; I ++ ) {Scanf (  "  % Lf  " , & GN [I]. X ,& GN [I]. Y );}  For ( Int I = 1 ; I <= m; I ++) {Scanf (  "  % Lf  " , & GM [I]. X ,& GM [I]. Y );}  Int CNT = 0  ;  For ( Int I = 1 ; I <= N; I ++ )  For ( Int J =1 ; J <= m; j ++ ) {G [CNT ++] = Cal (GN [I], GM [J]);} Sort (G, G + CNT );  Int B = 0 , D = CNT- 1  ;  While (B < D ){  Int Mid = (B + d )/ 2 ; // Because D is selected, D must not be equal to mid.              If (DLX (G [Mid]) = 1  ) {D = Mid; //  The key thing to grasp is to find the largest and the smallest, so that which of the = mid is not equal to the mid as much as possible.  }  Else  {B = Mid + 1  ;}}  Double  Ans; ans = (G [B] *10000000 + 0.5 )/ 10000000  ; Printf (  "  %. 6lf \ n  "  , G [B]);}  Return   0  ;}