SSL P2645 线段树练习题二

题目大意：
N长度的桌子上零散地放着M个不同颜色的盒子，桌子的后方是一堵墙。给出每个箱子的左端跟右端，问从桌子前方可以看到多少个盒子？假设人站得足够远(输入时,由底向上,从左到右)。

1<=n<=100000,1<=m<=100000,保证坐标[l,r]范围为[1,n].
题解：
线段树：
跟ZJU的那道题一样，
http://blog.csdn.net/gx_man_vip/article/details/70160195
不过把记录数量变成了记录是否存在。

var
      ans,tree:array [0..300000] of longint;
      sum,i,n,m,a,b,c:longint;

procedure insert(p,l,r,a,b:longint);
var
      mid:longint;
begin
      mid:=(l+r) div 2;
      if (a=l) and (b=r)
         then tree[p]:=c
         else begin
                  if tree[p]<>-1
                     then begin
                               tree[p * 2]:=tree[p];
                               tree[p*2+1]:=tree[p];
                               tree[p]:=-1;
                          end;
                  if b<=mid then insert(p*2,l,mid,a,b)
                            else if a>=mid then insert(p*2+1,mid,r,a,b)
                                           else begin
                                                     insert(p * 2,l,mid,a,mid);
                                                     insert(p*2+1,mid,r,mid,b);
                                                end;
              end;
end;

procedure count(p,l,r:longint);
var
      mid:longint;
begin
      mid:=(l+r) div 2;
      if tree[p]=-2
         then begin
                  c:=-2;
                  exit;
              end
         else if (tree[p]<>c) and (tree[p]<>-1)
                 then begin
                            c:=tree[p];
                            ans[c]:=1;
                      end
                 else if (r-l=1) and (c=tree[p])
                         then exit
                         else if tree[p]=-1
                                 then begin
                                           count(p * 2,l,mid);
                                           count(p*2+1,mid,r);
                                      end;

end;

procedure print;
begin
     for i:=1 to n do
         if ans[i]=1 then inc(sum);
     writeln(sum);
end;

begin
           for i:=0 to 300000 do tree[i]:=-2;
           readln(n);
           readln(m);
           for i:=1 to m do
           begin
                readln(a,b);
                c:=i;
                insert(1,1,n,a,b);
           end;
           c:=-3;
           count(1,1,n);
           print;
end.