poj1823,3667

又来练线段树了……

poj1823题意很简单（明显的数据结构题），区间修改和统计最长连续空区间；

有了poj3468的基础，区间修改不是什么问题了，重点是求最长连续空区间；（弱弱的我纠结了好久）

在每个节点上增加3个域，lmax，rmax，maxx，分别表示从左起最长，右起最长，区间内最长，然后稍稍动点脑筋即可……

 type node=record
        lmax,rmax,maxx:longint;
        l,r,lazy:integer;   //lazy表示区间三种情况，-表示区间内有人住但未满，表示无人住，表示全住满
      end;
 var tree:array[..] of node;
     i,n,m,a,b,c,j:longint;
 procedure updata(i:longint);     //更新
   var p:longint;
   begin
     if tree[i].lazy<>- then
     begin
       if tree[i].lazy= then p:=tree[i].r-tree[i].l+ else p:=;
       tree[i].lmax:=p;
       tree[i].rmax:=p;
       tree[i].maxx:=p;
     end
     else begin        //看起来很繁琐，实际上仔细想想就明白了
       tree[i].lmax:=tree[i*].lmax;
       if tree[i].lmax=tree[i*].r-tree[i*].l+ then tree[i].lmax:=tree[i].lmax+tree[i*+].lmax;
       tree[i].rmax:=tree[i*+].rmax;
       if tree[i].rmax=tree[i*+].r-tree[i*+].l+ then tree[i].rmax:=tree[i].rmax+tree[i*].rmax;
       p:=max(tree[i].lmax,tree[i].rmax);
       p:=max(p,max(tree[i*].maxx,tree[i*+].maxx));
       tree[i].maxx:=max(p,tree[i*].rmax+tree[i*+].lmax);
     end;
   end;

 procedure pushdown(i:longint);  //lazy思想，注意标记下移的时候不忘更新，因为某个子区间不一定会被访问
   begin
     tree[i*].lazy:=tree[i].lazy;
     updata(i*);
     tree[i*+].lazy:=tree[i].lazy;
     updata(i*+);
     tree[i].lazy:=-;    //需要标记下移的时候当前区间一定不会是全满或全空（想想为什么）
   end;

 procedure build(i,l,r:longint);  //初始化线段树
   var m:longint;
   begin
     tree[i].l:=l; tree[i].r:=r;
     tree[i].lazy:=;
     updata(i);
     if l<>r then
     begin
       m:=(l+r) div ;
       build(i*,l,m);
       build(i*+,m+,r);
     end;
   end;

 procedure work(i,l,r,t:longint);
   var m:longint;
   begin
     if (l>=a) and (r<=b) then
     begin
       tree[i].lazy:=t;
       updata(i);
     end
     else if l<>r then begin
       if tree[i].lazy<>- then pushdown(i); 
       m:=(l+r) div ;
       if (a<=m) then work(i*,l,m,t);
       if (b>=m+) then work(i*+,m+,r,t);
       updata(i);   // 左右子区间访问过后更新当前区间
     end;
   end;
 begin
   readln(n,m);
   build(,,n);
   for i:= to m do
   begin
     read(c);
     if c<> then
     begin
       readln(a,b);
       b:=a+b-;
       if c= then work(,,n,) else work(,,n,);  //代表入住，代表退房
     end
     else if c= then writeln(tree[].maxx);
   end;
 end.

poj1823

话说线段树查起来真累（难道还是我太渣了？），耗了5次才AC

poj3667在poj1823基础上多了个查找但并不难，对于当前区间，按照从左区间到右区间的优先顺序找即可，一次AC

 type node=record
        l,r,lmax,rmax,maxx:longint;
        lazy:integer;
      end;

 var tree:array[..] of node;
     i,n,m,a,b,c,j,l:longint;
 function max(a,b:longint):longint;
   begin
     if a>b then max:=a else max:=b;
   end;

  function min(a,b:longint):longint;
    begin
      if a= then exit(b);
      if b= then exit(a);
      if a>b then exit(b) else exit(a);
    end;
 procedure updata(i:longint);
   var p:longint;
   begin
     if tree[i].lazy<>- then
     begin
       if tree[i].lazy= then p:=tree[i].r-tree[i].l+ else p:=;
       tree[i].lmax:=p;
       tree[i].rmax:=p;
       tree[i].maxx:=p;
     end
     else begin
       tree[i].lmax:=tree[i*].lmax;
       if tree[i].lmax=tree[i*].r-tree[i*].l+ then tree[i].lmax:=tree[i].lmax+tree[i*+].lmax;
       tree[i].rmax:=tree[i*+].rmax;
       if tree[i].rmax=tree[i*+].r-tree[i*+].l+ then tree[i].rmax:=tree[i].rmax+tree[i*].rmax;
       p:=max(tree[i].lmax,tree[i].rmax);
       p:=max(p,max(tree[i*].maxx,tree[i*+].maxx));
       tree[i].maxx:=max(p,tree[i*].rmax+tree[i*+].lmax);
     end;
   end;

 procedure pushdown(i:longint);
   begin
     tree[i*].lazy:=tree[i].lazy;
     updata(i*);
     tree[i*+].lazy:=tree[i].lazy;
     updata(i*+);
     tree[i].lazy:=-;
   end;

 procedure find(i:longint);
   begin
     if tree[i].maxx<l then exit;
     if (tree[i].lmax>=l) then a:=tree[i].l
     else if tree[*i].maxx>=l then
       find(*i)
     else if tree[i*].rmax+tree[i*+].lmax>=l  then
     begin
       a:=tree[i*].r-tree[i*].rmax+;
     end
     else if tree[i*+].maxx>=l then find(*i+);
     if tree[i].rmax>=l then a:=min(a,tree[i].r-tree[i].rmax+);
   end;

 procedure build(i,l,r:longint);
   var m:longint;
   begin
     tree[i].l:=l; tree[i].r:=r;
     tree[i].lazy:=;
     updata(i);
     if l<>r then
     begin
       m:=(l+r) div ;
       build(i*,l,m);
       build(i*+,m+,r);
     end;
   end;

 procedure work(i,l,r,t:longint);
   var m:longint;
   begin
     if (l>=a) and (r<=b) then
     begin
       tree[i].lazy:=t;
       updata(i);
     end
     else if l<>r then begin
       if tree[i].lazy<>- then pushdown(i);
       m:=(l+r) div ;
       if (a<=m) then work(i*,l,m,t);
       if (b>=m+) then work(i*+,m+,r,t);
       updata(i);
     end;
   end;

 begin
   readln(n,m);
   build(,,n);
   for i:= to m do
   begin
     read(c);
     if c= then
     begin
       readln(l);
       a:=;
       find();
       writeln(a);
       b:=a+l-;
       if a<> then work(,,n,);
     end
     else if c= then
     begin
       readln(a,b);
       b:=a+b-;
       work(,,n,);
     end;
   end;
 end.

poj3667