poj 2528 Mayor's posters 线段树

来源：互联网发布：淘宝汽车冬季坐垫编辑：程序博客网时间：2024/05/29 08:33

题意：n（n<=10000)个人依次贴海报,给出每张海报所贴的范围li，ri（1<=li<=ri<=10000000)。求出最后还能看见多少张海报。

分析：一看就知道是线段树啦。然后就先离散化一下，再就是个普通的线段树啦。注意在插入操作时要把原来的海报覆盖掉就好啦。（啦够了没有）

代码：

var  o,e,i,j,ans,x,y,n:longint;  l,r:array[1..10000] of longint;  a:array[0..20000,1..2] of longint;  v:array[1..10000] of boolean;  t:array[1..400000,1..3] of longint;procedure qsort(l,r:longint);var  i,j,k:longint;begin  if l>=r then exit;  i:=l;  j:=r;  k:=a[(i+j) div 2,1];  repeat    while a[i,1]<k do inc(i);    while a[j,1]>k do dec(j);    if i<=j then    begin      a[0]:=a[i];a[i]:=a[j];a[j]:=a[0];      inc(i);dec(j);    end;  until i>j;  qsort(i,r);  qsort(l,j);end;procedure hehe(d,l,r:longint);var  m:longint;begin  t[d,1]:=l;  t[d,2]:=r;  t[d,3]:=0;  if l=r then exit;  m:=(l+r) div 2;  hehe(d*2,l,m);  hehe(d*2+1,m+1,r);end;procedure insert(d,l,r,z:longint);var  m:longint;begin  if (t[d,1]=l)and(t[d,2]=r) then  begin    t[d,3]:=z;    exit;  end;  if t[d,3]>0 then  begin    t[d*2,3]:=t[d,3];    t[d*2+1,3]:=t[d,3];    t[d,3]:=0;  end;  m:=(t[d,1]+t[d,2]) div 2;  if r<=m    then insert(d*2,l,r,z)    else if l>m           then insert(d*2+1,l,r,z)           else begin                  insert(d*2,l,m,z);                  insert(d*2+1,m+1,r,z);                end;end;procedure count(d:longint);begin  if t[d,3]>0 then  begin    v[t[d,3]]:=false;    exit;  end;  if t[d,1]=t[d,2] then exit;  count(d*2);  count(d*2+1);end;begin  readln(o);  for e:=1 to o do  begin    readln(n);    for i:=1 to n do    begin      readln(x,y);      a[i*2-1,1]:=x;      a[i*2-1,2]:=i;      a[i*2,1]:=y;      a[i*2,2]:=i;    end;    qsort(1,n*2);    j:=0;    a[0,1]:=0;    fillchar(l,sizeof(l),0);    fillchar(r,sizeof(r),0);    for i:=1 to n*2 do    begin      if a[i,1]<>a[i-1,1] then inc(j);      if l[a[i,2]]=0       then l[a[i,2]]:=j       else r[a[i,2]]:=j;    end;    hehe(1,1,j);    for i:=1 to n do      insert(1,l[i],r[i],i);    fillchar(v,sizeof(v),true);    count(1);    ans:=0;    for i:=1 to n do      if v[i]=false then inc(ans);    writeln(ans);  end;end.

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