线段树初步

线段树模板1：https://www.luogu.org/problem/show?pid=3372

线段树模板2：https://www.luogu.org/problem/show?pid=3373

这些都比较基础，就是1或2个lazy标记的时候怎么处理？几乎不用考虑兼容性的问题。

现在这里有一道充分考验线段树lazy的兼容性问题的题目，涉及到4个lazy标记，怎么处理？

例子1：求线段树维护一个区间，支持如下操作：区间修改为同一个值（更改1），区间加一个数（更改2），区间和（运算1），区间最大值（运算2）：

解析：首先，不考虑lazy标记的兼容性是不行的，需要注意的是区间修改和区间和，区间最大值是不兼容的，所以在lazy时需要特判！！！

还有需要注意的点非常多，需要注意，

//本程序经过oycy0306测试正确性保障++uses math;
const maxn=1000;
      inf=233333333;
type rec=record
add,mk:longint;
end;
var  n,m,i,opx,opr,opl,ch,ans:longint;
     f,ff,a:array[1..maxn]of longint;
     s:array[1..maxn]of rec;
procedure build(x,l,r:longint);
var m:longint;
begin
 s[x].mk:=inf;
 s[x].add:=0;
 if l=r then begin
  ff[x]:=a[l];
  f[x]:=a[l];
  exit;
 end;
 m:=(l+r)>>1;
 build(2*x,l,m);
 build(2*x+1,m+1,r);
 f[x]:=f[2*x]+f[2*x+1];
 ff[x]:=max(ff[2*x],ff[2*x+1]);
end;
procedure down(x,l,r,m,lson,rson:longint);
begin
 if s[x].mk<>inf then begin  //**
  s[lson].mk:=s[x].mk;
  s[rson].mk:=s[x].mk;
  s[lson].add:=0;
  s[rson].add:=0;
  f[lson]:=opx*(m-l+1);
  f[rson]:=opx*(r-m);
  ff[lson]:=opx;
  ff[rson]:=opx;
  s[x].mk:=inf;
  s[x].add:=0;
  exit;
 end;
  s[lson].mk:=inf;
  s[rson].mk:=inf;
  s[lson].add:=s[lson].add+s[x].add;
  s[rson].add:=s[rson].add+s[x].add;
  f[lson]:=f[lson]+s[x].add*(m-l+1);
  f[rson]:=f[rson]+s[x].add*(r-m);
  ff[lson]:=ff[lson]+s[x].add;
  ff[rson]:=ff[lson]+s[x].add;
  s[x].add:=0;
  s[x].mk:=inf;             //**
end;
procedure calc(x,l,r:longint);
var m:longint;
begin
 if  (opl<=l)and(opr>=r) then begin
  case ch of
   1:begin s[x].mk:=opx; s[x].add:=0; f[x]:=opx*(r-l+1); ff[x]:=opx; end;
   2:begin s[x].add:=s[x].add+opx; f[x]:=f[x]+opx*(r-l+1); ff[x]:=ff[x]+opx; end;
   3:begin ans:=ans+f[x]; end;  //f:sum
   4:begin ans:=max(ff[x],ans)end;  //ff:mk
  end;
  exit;
  end;
  m:=(l+r)>>1;
  if (s[x].mk<>inf)or(s[x].add>0)then down(x,l,r,m,2*x,2*x+1);
  if opl<=m then calc(2*x,l,m);
  if opr>m then calc(2*x+1,m+1,r);
  if (ch=1)or(ch=2) then begin
   f[x]:=f[2*x]+f[2*x+1];
   ff[x]:=max(ff[2*x],ff[2*x+1])
  end;
end;
begin
 writeln(‘input nodenum n=??‘); readln(n);
 writeln(‘input a num(n) sequence called a[]==??‘);
 for i:=1 to n do read(a[i]);
 write(‘the strat sequence a[]==‘);
 for i:=1 to n do write(a[i],‘ ‘);writeln;
 writeln(‘---------------------------‘);
 writeln(‘start to build XD tree.‘);
 writeln(‘---------------------------‘);
 build(1,1,n);
 writeln(‘---------------------------‘);
 writeln(‘XD tree is already built.‘);
 writeln(‘---------------------------‘);
 writeln(‘input your instructions number!‘); //build ok!
 readln(m); writeln(‘OK.‘);
 writeln(‘---------------------------‘);
 writeln(‘1=modify‘);
 writeln(‘2=ADD‘);
 writeln(‘3=question sum‘);
 writeln(‘4=max in the a[]‘);
 writeln(‘a line a word.‘);
 writeln(‘instruction+ +minl+ +maxr+ (+valuable)‘);
 writeln(‘---------------------------‘);
 for i:=1 to m do begin
  writeln(‘instruction input ‘,i,‘ :‘);
  read(ch,opl,opr);
  if (ch=3)or(ch=4) then readln else readln(opx);
  case ch of
   1,2:begin writeln(‘instruction output ‘,i,‘:‘); writeln(‘Nothing except the instruction is running over.‘);calc(1,1,n); end;
   3,4:begin ans:=0; calc(1,1,n); writeln(‘instruction output ‘,i,‘ :‘); writeln(ans); end;
  end;
 end;
end.

 例子2：求线段树维护一个区间，支持如下操作：区间修改为同一个值（更改1），区间加一个数（更改2），区间乘一个数（更该3），区间和（运算1），区间最大值（运算2）：、

对于例子1又多了一个区间乘法，所以就有5个lazy标记了...

所以这个down函数比较的长； 具体不解释了，就是在例子1上增加，看代码：

uses math;
const maxn=1000;
      inf=233333333;
type rec=record
add,mk,mp:longint;
end;
var  n,m,i,opx,opr,opl,ch,ans:longint;
     f,ff,a:array[1..maxn]of longint;
     s:array[1..maxn]of rec;
procedure build(x,l,r:longint);
var m:longint;
begin
 s[x].mk:=inf;
 s[x].add:=0;
 s[x].mp:=1;
 if l=r then begin
  ff[x]:=a[l];
  f[x]:=a[l];
//  s[x].mx:=a[l];
  exit;
 end;
 m:=(l+r)>>1;
 build(2*x,l,m);
 build(2*x+1,m+1,r);
 f[x]:=f[2*x]+f[2*x+1];
 ff[x]:=max(ff[2*x],ff[2*x+1]);
end;
procedure down(x,l,r,m,lson,rson:longint);
begin
 if s[x].mk<>inf then begin  //**
  s[lson].mk:=s[x].mk;
  s[rson].mk:=s[x].mk;
  s[lson].add:=0;
  s[rson].add:=0;
  s[lson].mp:=1;
  s[rson].mp:=1;
  f[lson]:=opx*(m-l+1);
  f[rson]:=opx*(r-m);
  ff[lson]:=opx;
  ff[rson]:=opx;
  s[x].mk:=inf;
  s[x].add:=0;
  s[x].mp:=1;
  exit;
 end;
  s[lson].mp:=s[lson].mp*s[x].mp;
  s[rson].mp:=s[rson].mp*s[x].mp;
  s[lson].add:=s[lson].add*s[x].mp;
  s[rson].add:=s[rson].add*s[x].mp;
  f[lson]:=f[lson]*s[x].mp;
  f[rson]:=f[rson]*s[x].mp;
  ff[lson]:=ff[lson]*s[x].mp;
  ff[rson]:=ff[lson]*s[x].mp;
  s[x].mp:=1;
  s[lson].mk:=inf;
  s[rson].mk:=inf;
  s[lson].add:=s[lson].add+s[x].add;
  s[rson].add:=s[rson].add+s[x].add;
  f[lson]:=f[lson]+s[x].add*(m-l+1);
  f[rson]:=f[rson]+s[x].add*(r-m);
  ff[lson]:=ff[lson]+s[x].add;
  ff[rson]:=ff[lson]+s[x].add;
  s[x].add:=0;
  s[x].mk:=inf;             //**
end;
procedure calc(x,l,r:longint);
var m:longint;
begin
 if  (opl<=l)and(opr>=r) then begin
  case ch of
   1:begin s[x].mk:=opx; s[x].add:=0; f[x]:=opx*(r-l+1); ff[x]:=opx; end;
   2:begin s[x].add:=s[x].add+opx; f[x]:=f[x]+opx*(r-l+1); ff[x]:=ff[x]+opx; end;
   3:begin ans:=ans+f[x]; end;  //f:sum
   4:begin ans:=max(ff[x],ans)end;  //ff:mk
   5:begin s[x].mp:=s[x].mp*opx; s[x].add:=s[x].add*opx; f[x]:=f[x]*opx; ff[x]:=ff[x]*opx; end;
  end;
  exit;
  end;
  m:=(l+r)>>1;
  if (s[x].mk<>inf)or(s[x].add>0)or(s[x].mp<>1)then down(x,l,r,m,2*x,2*x+1);
  if opl<=m then calc(2*x,l,m);
  if opr>m then calc(2*x+1,m+1,r);
  if (ch=1)or(ch=2) then begin
   f[x]:=f[2*x]+f[2*x+1];
   ff[x]:=max(ff[2*x],ff[2*x+1])
  end;
end;
begin
 writeln(‘input nodenum n=??‘); readln(n);
 writeln(‘input a num(n) sequence called a[]==??‘);
 for i:=1 to n do read(a[i]);
 write(‘the strat sequence a[]==‘);
 for i:=1 to n do write(a[i],‘ ‘);writeln;
 writeln(‘---------------------------‘);
 writeln(‘start to build XD tree.‘);
 writeln(‘---------------------------‘);
 build(1,1,n);
 writeln(‘---------------------------‘);
 writeln(‘XD tree is already built.‘);
 writeln(‘---------------------------‘);
 writeln(‘input your instructions number!‘); //build ok!
 readln(m); writeln(‘OK.‘);
 writeln(‘---------------------------‘);
 writeln(‘1=modify‘);
 writeln(‘2=ADD‘);
 writeln(‘3=question sum‘);
 writeln(‘4=max in the a[]‘);
 writeln(‘5=multiplication‘);
 writeln(‘a line a word.‘);
 writeln(‘instruction+ +minl+ +maxr+ (+valuable)‘);
 writeln(‘---------------------------‘);
 for i:=1 to m do begin
  writeln(‘instruction input ‘,i,‘ :‘);
  read(ch,opl,opr);
  if (ch=3)or(ch=4) then readln else readln(opx);
  case ch of
   1,2,5:begin calc(1,1,n); writeln(‘instruction output ‘,i,‘:‘); writeln(‘Nothing except the instruction is running over.‘); end;
   3,4:begin ans:=0; calc(1,1,n); writeln(‘instruction output ‘,i,‘ :‘); writeln(ans); end;
  end;
 end;
end.