poj 1823 hotel（线段树）

题目

线段树的区间更新

Problem: 1823       Memory: 1644K       Time: 2938MSprogram hotel;const maxN=16000*4;type rec = record      left,right,value,lm,rm,mark:longint; end;     //lm表示从左开始最多连续多少个，lr同理var                                                       n,k,i,a,mem,l:longint;  tree:array[1..maxN]of rec;function max(a,b:longint):longint;begin    if a>b then max:=a    else max:=b;end;procedure buildtree(l,r,k:longint);var mid:longint;begin tree[k].left:=l; tree[k].right:=r; tree[k].value:=r-l+1; tree[k].lm:=r-l+1; tree[k].rm:=r-l+1; tree[k].mark:=-1;  if l=r then exit; mid:=(l+r)div 2; buildtree(l,mid,k*2); if mid<r then  buildtree(mid+1,r,k*2+1);end;procedure pushdown(k:longint);var t:longint;begin if tree[k].mark=-1 then exit; //-1表示已经更新完毕 if tree[k].mark=0 then begin  t:=tree[k*2].right-tree[k*2].left+1;  tree[k*2].value:=t;  tree[k*2].lm:=t;  tree[k*2].rm:=t;  tree[k*2].mark:=0;  t:=tree[k*2+1].right-tree[k*2+1].left+1;  tree[k*2+1].value:=t;  tree[k*2+1].lm:=t;  tree[k*2+1].rm:=t;  tree[k*2+1].mark:=0; end else begin  tree[k*2].value:=0;  tree[k*2].lm:=0;  tree[k*2].rm:=0;  tree[k*2].mark:=1;  tree[k*2+1].value:=0;  tree[k*2+1].lm:=0;  tree[k*2+1].rm:=0;  tree[k*2+1].mark:=1; end; tree[k].mark:=-1;end;procedure pushup(k:longint);var a,b,c:longint;begin a:=tree[k*2].value; b:=tree[k*2+1].value; c:=tree[k*2].rm+tree[k*2+1].lm; tree[k].value:=max(max(a,b),c); tree[k].lm:=tree[k*2].lm; tree[k].rm:=tree[k*2+1].rm; if (tree[k*2].lm=(tree[k*2].right-tree[k*2].left+1)) then tree[k].lm:=tree[k].lm+tree[k*2+1].lm; if (tree[k*2+1].rm=(tree[k*2+1].right-tree[k*2+1].left+1)) then tree[k].rm:=tree[k].rm+tree[k*2].rm;end;procedure update(l,r,w,k:longint);begin if (tree[k].left=l)and(tree[k].right=r) then begin   if w=1 then begin    tree[k].value:=0;    tree[k].mark:=1;    tree[k].lm:=0;    tree[k].rm:=0;   end   else if w=0 then begin    tree[k].value:=r-l+1;    tree[k].mark:=0;    tree[k].lm:=r-l+1;    tree[k].rm:=r-l+1;   end;  exit; end; pushdown(k);//更新子树 if tree[k*2].right>=r then   update(l,r,w,k*2) else if tree[k*2+1].left<=l then update(l,r,w,k*2+1) else begin  update(l,tree[k*2].right,w,k*2);  update(tree[k*2+1].left,r,w,k*2+1); end; pushup(k);//更新父节点值域end;begin readln(n,k); buildtree(1,n,1); for i:=1 to k do begin  read(a);  if a=3 then writeln(tree[1].value)  else begin   readln(l,mem);   if a=1 then update(l,l+mem-1,1,1)   else update(l,l+mem-1,0,1);  end; end;end.

反思：
1.发现自己的分析能力还有很大上升空间…看其他人写题解什么的能把问题分析得清清楚楚的，我完全不知道怎么写题解…还是练习少了吧…
2.觉得代码还有可以简化的地方，继续加油