Cow Crossings

Posted 2021-01-10 godrose

题面：

思路：

       求非逆序对的个数

代码：

var n,i,j,ans,k:longint;
 a,b:Array[0..100005] of longint;
  f:array[0..100005] of boolean;

    procedure sort(l,r: longint);
      var
         i,j,x,y: longint;
      begin
         i:=l;
         j:=r;
         x:=b[(l+r) div 2];
         repeat
           while b[i]<x do
            inc(i);
           while x<b[j] do
            dec(j);
           if not(i>j) then
             begin
                y:=a[i];
                a[i]:=a[j];
                a[j]:=y;
                y:=b[i];
                b[i]:=b[j];
                b[j]:=y;
                inc(i);
                j:=j-1;
             end;
         until i>j;
         if l<j then
           sort(l,j);
         if i<r then
           sort(i,r);
      end;


begin

assign(input,‘crossings.in‘);
reset(input);
assign(output,‘crossings.out‘);
rewrite(output);

 read(n);
 for i:=1 to n do read(a[i],b[i]);

 sort(1,n);

 k:=-maxlongint;
 for i:=1 to n do
  if k>a[i] then f[i]:=true else k:=a[i];

 k:=maxlongint;
 for i:=n downto 1 do
  if k<a[i] then f[i]:=true else k:=a[i];

 ans:=0;

 for i:=1 to n do
  if f[i] then inc(ans);

 writeln(n-ans);


close(input);close(output);
end.