PKU 3275 Ranking the Cows 最短路 floyd

题意/Description：

    Each of Farmer John's N cows (1 ≤ N ≤ 1,000) produces milk at a different positive rate, and FJ would like to order his cows according to these rates from the fastest milk producer to the slowest.

    FJ has already compared the milk output rate for M (1 ≤ M ≤ 10,000) pairs of cows. He wants to make a list of C additional pairs of cows such that, if he now compares those C pairs, he will definitely be able to deduce the correct ordering of all N cows. Please help him determine the minimum value of C for which such a list is possible.

 

读入/Input：

    Line 1: Two space-separated integers: N and M 
    Lines 2..M+1: Two space-separated integers, respectively: X and Y. Both X and Y are in the range 1...N and describe a comparison where cow X was ranked higher than cow Y.

 

输出/Output：

    Line 1: A single integer that is the minimum value of C.

 

题解/solution：

   如果n头牛排序完成，应该有C(n,2)关系已知，即任意两头牛的速度都知道了。然后可以从已经比较了的m对牛中算出可以推导出多少对牛已经知道了，推导方法可以参考floyd最短路，然后用C（n,2）减去它就是答案。

   补充：C(N, 2) = N * (N - 1) / 2

   调查K对关系：C(N, 2) - K

 

代码/Code：

var
  a,b:array [0..10001,0..1001] of longint;
  f:array [0..1001,0..1001] of boolean;
  n,m,ans:longint;
procedure init;
var
  i,x,y:longint;
begin
  readln(n,m);
  for i:=1 to m do
    begin
      readln(x,y);
      f[x,y]:=true;
      inc(a[x,0]); inc(b[y,0]);
      a[x,a[x,0]]:=y; b[y,b[y,0]]:=x;
    end;
  ans:=0;
end;

procedure main;
var
  i,j,k,x,y:longint;
begin
  for k:=1 to n do
    for i:=1 to b[k,0] do
      begin
        x:=b[k,i];
        for j:=1 to a[k,0] do
          begin
            y:=a[k,j];
            if not f[x,y] then
              begin
                f[x,y]:=true;
                inc(a[x,0]); inc(b[y,0]);
                a[x,a[x,0]]:=y; b[y,b[y,0]]:=x;
                inc(ans);
              end;
          end;
      end;
  write(n*(n-1) div 2-ans-m);
end;

begin
  init;
  main;
end.