Connect the server through the IP address and port



The gopher family, having averted the canine threat, must face a new predator. 

The are n gophers and m gopher holes, each at distinct (x, y) coordinates. A hawk arrives and if a gopher does not reach a hole in s seconds it is vulnerable to being eaten. A hole can save at most one gopher. All the gophers run at the same velocity v. The gopher family needs an escape strategy that minimizes the number of vulnerable gophers.


The input contains several cases. The first line of each case contains four positive integers less than 100: n, m, s, and v. The next n lines give the coordinates of the gophers; the following m lines give the coordinates of the gopher holes. All distances are in metres; all times are in seconds; all velocities are in metres per second.


Output consists of a single line for each case, giving the number of vulnerable gophers.

Sample input

2 2 5 10
1.0 1.0
2.0 2.0
100.0 100.0

20.0 20.0

Sample output


This question is relatively intended … It is said that there are N mice, M holes, the time S of the eagle reaches the ground, the mouse’s movement speed V, how many pickers will be eaten by the eagle. If this is established, the mouse and hole should be regarded as two sets. Then only the time from the local mouse to the cave is less than the time from the eagle to the ground.

#include <stdio.h> 
 #include <string.h> 
 #include <string> 
 #include <math.h> 
 #include <vector> 
 #include <queue> 
 #include <stack> 
 #include <map> 
 #include <iOSTREAM> 
 #include <algorithm> 
 using namespace std; 
 int n, m, ma [505] [505], vis [505], dx [505]; 
 struct point 
     double x, y; 
 } G [200], h [200]; 
 Double DIS (Point P1, Point P2) 
 {// The time of calculating the mouse to the hole 
     Return SQRT ((P2.X-P1.x)*(P2.x-P1.x)+(P2.y-P1.y)*(p2.y-p1.y)); 
 int Find (int i) 
 for (int j = n+1; j <= n+m; j ++) 
 if (ma [i] [j] && vis [j] == 0) 
 vis [j] = 1; 
 if (dx [j] ==-1 || Find (dx [j]))) 
 dx [j] = i; 
 Return 1; 
 Return 0; 
 int Main () 
 int T, I, J, S, V, X, Y, Sum; 
 While (Scanf ("%D%D%D%D", & N, & M, & S, & V)! = EOF) 
 if (t == 0) Break; 
 Memset (ma, 0, sizeof ma); 
 Memset (dx, -1, sizeof dx); 
 for (i = 1; i <= n; i ++) 
         scanf ("%lf%lf", & g [i] .x, & g [i] .y); 
         for (i = 1; i <= m; i ++) 
         scanf ("%lf%lf", & h [i] .x, & h [i] .y); 
         for (i = 1; i <= n; i ++) 
             for (j = 1; j <= m; j ++) 
                 double d = dis (g [i], h [j]); 
                 if (d/v <= (double) s) // is safer than the eagle 
                 ma [i] [n+j] = 1; 
 sum = 0; 
 for (i = 1; i <= n; i ++) 
 SUM+= Find (i); 
 Printf ("%d \ n", n-sum); 


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