Pku2060 taxi cab Scheme

Question:

A taxi can keep running its tasks. The task is to drive a taxi from one place to another within a certain period of time. Give the specific time and location of each task, and ask at leastHow many taxis are used for the task?

Analysis:

First of all, because we are doing binary matching, we must know that we are using binary matching. If it happens by accident, I am not sure I can think of binary matching;

Since binary matching is used, the first part is the composition. How can we establish a matching relationship? It is to establish a relationship between tasks and regard any tasks that may be joined together as a match. Obviously, the next step is to find a minimum path overwrite in this figure, the minimum path overwrites all vertices to complete all tasks;

Minimum path overwrite = number of vertices-maximum match

In the calculation of the time, all the time is converted into a component, which reduces a lot of judgment.

# Include <iostream> # include <math. h> using namespace STD; bool map [501] [501], vis [501]; int match [501], n; struct road {int m; int EI, EJ ;} R [501]; int path (INT s) {for (INT I = 1; I <= N; I ++) if (Map [s] [I] &! Vis [I]) {vis [I] = 1; if (Match [I] =-1 | path (Match [I]) {match [I] = s; return 1 ;}} return 0 ;}int max_match () {memset (match,-1, sizeof (MATCH); int ans = 0; for (INT I = 1; I <= N; I ++) {memset (VIS, 0, sizeof (VIS); ans + = path (I );} return ans;} int main () {int cas, A, B, C, D, H1, M1, temp; scanf ("% d", & CAS ); while (CAS --) {scanf ("% d", & N); memset (MAP, 0, sizeof (MAP); For (INT I = 1; I <= N; I ++) {scanf ("% d: % d", & H1, & M1, & A, & B, & C, & D); R [I]. ei = C; R [I]. j = D; temp = ABS (c-A) + ABS (B-d); R [I]. M = temp + M1 + H1 * 60; For (Int J = 1; j <I; j ++) {temp = ABS (A-R [J]. EI) + ABS (B-r [J]. j); If (R [J]. m + temp <M1 + H1 * 60) map [I] [J] = 1 ;}} printf ("% d \ n", n-Max_Match ());} return 0 ;}
 

I still don't know why this is so much faster than me.

Pku2060 accelerated version

 # Include  <  Iostream  >  

  Using     Namespace  STD;

  Struct  Time
{
  Int  H, M;
};
  Struct  Task
{
  Int SX, Sy, ex, ey;
Time st;
};

Task V [  505  ];

  Char  Mat [  505  ] [  505  ];
  Int  MX [  505  ], My [  505  ];
  Char  Vstd [ 505  ];
  Int  NV;


  Bool  Early (Task A, Task B)
{
  Int  T  =  ABS (A. Ex  -  B. SX)  +  ABS (A. ey  -  B. Sy );
T  + = ABS (A. SX  -  A. Ex)  +  ABS (A. Sy  -  A. Ey );

A. St. m  + =  T;
A. St. h  + =  A. St. m  /  60  ;
A. St. m  % =     60 ;

  If  (A. St. h  <  B. St. h)
{
  Return     True  ;
}
  Else     If  (A. St. h  =  B. St. h)
{
  If  (A. St. m <  B. St. m)
  Return     True  ;
}

  Return     False  ;
}

  Int  PATH (  Int  S)
{
  Int  E;
  For (E  =     1  ; E  <=  NV; e  ++  )
{
  If  (MAT [s] [E]  &&     !  Vstd [e])
{
Vstd [E]  =    1  ;
  If  (My [E]  =-  1     |  PATH (my [e])
{
My [E]  =  S;
MX [s]  =  E;
  Return     1 ;
}
}
}
  Return     0  ;
}

  Int  Maxmatch ()
{
  Int  Res  =     0  ;
Memset (MX,  -  1  , Sizeof  (MX ));
Memset (my,  -  1  ,  Sizeof  (My ));

  For  (  Int  I  =     1  ; I  <=  NV; I ++  )
{
  If  (MX [I]  =-  1  )
{
Memset (vstd,  0  ,  Sizeof  (Vstd ));
Res  + =  PATH (I );
}
}
  Return  Res;
}
  Int Main ()
{
  Int  CAS;
  Int  I, J;
Scanf (  "  % D  "  ,  &  CAS );
  While  (CAS  --  )
{
Scanf (  "  % D "  ,  &  NV );

  For  (I  =     1  ; I  <=  NV; I  ++  )
{
Scanf (  "  % D: % d  " ,  &  V [I]. St. H,  &  V [I]. St. m );
Scanf (  "  % D  "  ,  &  V [I]. SX,  &  V [I]. Sy,  &  V [I]. ex,  & V [I]. Ey );
}

Memset (mat,  0  ,  Sizeof  (MAT ));
  For  (I  =     1  ; I  <=  NV; I  ++  )
{
  For (J  =     1  ; J  <=  NV; j  ++  )
{
  If  (I  =  J)  Continue  ;
  If (Early (V [I], V [J])
{
Mat [I] [J]  =     1  ;
}
}
}

  Int  Res  =  NV  -  Maxmatch ();
Printf (  "  % D \ n  " , Res );
}
  Return     0  ;
}