Hdoj 4081 Qin Shi Huang's national road system

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4081

 

Solution report:

First, find the Minimum Spanning Tree, and record the tree. Use the low, and Pre arrays.

The DP array indicates the maximum edge. This information can be obtained when Prim is obtained.

And then enumerate any two points to repair the magic path. This path may be in the least generated tree or not.

If not, replace the longest path on the MST between the two points. DP Array

If the path directly replaced with MST is also a DP Array

View code

# Include <iostream> # Include < String . H> # Include <Stdio. h> # Include <Algorithm> # Include <Cmath> Using  Namespace  STD;  # Define Maxn 1110 # Define INF ~ 0u> 1 Double  Map [maxn] [maxn];  Double  People [maxn];  Double  Low [maxn];  Double Max ( Double A, Double  B ){  Return A> B? A: B ;}  Struct  Node0 {  Double  X;  Double  Y;} node [maxn];  Double  DP [maxn] [maxn];  Int  N;  Int  Visit [maxn];  Int  Pre [maxn];  Double SUM;  Double Len ( Double X1, Double Y1, Double X2, Double  Y2 ){  Return SQRT (x1-x2) * (x1-x2) + (y1-y2) * (Y1- Y2 ));}  Void  Init () {memset (people,  0 , Sizeof  (People )); For ( Int I = 0 ; I <= maxn; I ++ )  For ( Int J = 0 ; J <= maxn; j ++ ) Map [I] [J] = INF; scanf (  "  % D  " ,& N );  For (Int I = 1 ; I <= N; I ++ ) Scanf (  "  % Lf  " , & Node [I]. X, & node [I]. Y ,& People [I]);}  Void  Map (){  For ( Int I = 1 ; I <n; I ++ ){  For (Int J = I + 1 ; J <= N; j ++ ) {Map [I] [J] = Map [J] [I] = Len (node [I]. X, node [I]. Y, node [J]. X, node [J]. Y );}}}  Double  Prim (){  Double  Temp;  Int  ID;  Double Ans = 0  ; Memset (visit,  0 , Sizeof  (Visit); memset (DP,  0 , Sizeof  (DP ));  For ( Int I = 1 ; I <= N; I ++) Pre [I] = 1  ;  For ( Int I = 1 ; I <= maxn; I ++) low [I] = INF; visit [ 1 ] = 1  ;  For ( Int I = 2 ; I <= N; I ++ ) Low [I] = Map [ 1  ] [I];  For ( Int I = 2 ; I <= N; I ++ ) {Temp = INF; For ( Int J = 1 ; J <= N; j ++ )  If (! Visit [J] & low [J] < Temp) temp = Low [J], id = J; ans + = Temp, visit [ID] = 1  ;  For ( Int J = 1 ; J <= N; j ++ )  If (Visit [J] & ID! = J) DP [ID] [J] = DP [J] [ID] = Max (DP [pre [ID] [J], temp );  For ( Int J = 1 ; J <= N; j ++ )  If (Low [J]> map [ID] [J] &! Visit [J]) low [J] = Map [ID] [J], pre [J] = ID ;}  Return  Ans ;}  Double  Solve (){ Double Temp = 0.0  ;  Double Ans = 0.0  ;  For ( Int I = 1 ; I <n; I ++ )  For ( Int J = I + 1 ; J <= N; j ++ ) {Temp = (People [I] + people [J])/(Sum-DP [I] [J]); ans = Max (temp, ANS );}  Return  Ans ;}  Int  Main (){  Int  Test;  For (CIN> test; test -- ) {Init (); Map (); sum = Prim (); printf (  "  %. 2f \ n  " , Solve ());}  Return   0  ;}