NOIP前夕：noi.openjudge,Gopher II

Gopher II

总Time Limit:2000msMemory Limit:65536kB

Description

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.

Input

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

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

1

一道二分图的题
知道考试之前不能学新算法了，可是看见了自己不会的，还是忍不住学了一晚上
二分图最大匹配的匈牙利算法。
不断求增广路，每次求出就能使答案+1
题中只要 sqrt(sqr(mouse[i].x-hole[j].x)+sqr(mouse[i].y-hole[j].y))<=v*s
就在老鼠的集合中i的元素和洞的集合中的j的元素连一条边

CODE:

var n,m,s,v:longint;
    i,j,k:longint;
    pole:array[1..500]of record
                         x,y:real;
                         end;
    mouse:array[1..500]of record
                          x,y:real;
                          end;
    edge:array[1..200,1..200]of boolean;
    match:array[1..500]of longint;
    used:array[1..500]of boolean;
    ans:longint;

function dfs(x:longint):boolean;
         var i,j,k:longint;
         begin for i:=1 to m do
                   if (edge[x,i])and(used[i])
                      then begin used[i]:=false;
                                 if match[i]=0
                                    then begin match[i]:=x;
                                               exit(true);
                                         end
                                    else begin if dfs(match[i])
                                                  then begin match[i]:=x;
                                                             exit(true);
                                                       end;
                                         end;
                           end;
               exit(false);
         end;

begin
      while not eof do
      begin

      fillchar(mouse,sizeof(mouse),0);
      fillchar(pole,sizeof(pole),0);
      readln(n,m,s,v);
      for i:=1 to n do
          readln(mouse[i].x,mouse[i].y);
      for i:=1 to m do
          readln(pole[i].x,pole[i].y);
      fillchar(edge,sizeof(edge),false);
      for i:=1 to n do
          for j:=1 to m do
              if sqrt(sqr(mouse[i].x-pole[j].x)+sqr(mouse[i].y-pole[j].y))<=s*v
                 then edge

转载于:https://www.cnblogs.com/spiderKK/p/4928404.html