【ZJOI2017 Round2练习&BZOJ4826】D1T2 sf（主席树，单调栈）

题意：

思路：From http://blog.csdn.net/neither_nor/article/details/70211150

对每个点i，单调栈求出左边和右边第一个大于i的位置，记为l[i]和r[i]

那么(l[i],r[i])会产生p1的贡献

左端点为l[i]，右端点在[i+1,r-1]的点对都会产生p1的贡献

右端点为r[i]，左端点在[l+1,i-1]的点对都会产生p2的贡献

将点对看成平面上的点，横坐标左端点纵坐标右端点，上述贡献分别对应单点加和线段加

查询就是矩形求和

From MG：

考虑修改均为如(x,y1..y2)与(x1..x2,y)形式

因为询问均为矩形，则修改(y1..y2,x)与(x,y1..y2)等价

若统一转化为(x,y1..y2)形式，可用主席树维护

若统一转化为(y1..y2,x)形式，可用树状数组套主席树维护

注意点对(i,i)都有p1的贡献

 var t:array[..]of record
                            a,s:int64;
                            l,r:longint;
                           end;
     d:array[..,..]of longint;
     l,r:array[..]of longint;
     root,stk,a:array[..]of longint;
     n,m,i,cnt,j,top,que,x,y,p1,p2:longint;
     ans:int64;

 procedure update(l,r,x,y,v:longint;var p:longint);
 var mid:longint;
 begin
  if (l>x)or(x>y) then exit;
  inc(cnt); t[cnt]:=t[p]; p:=cnt;
  t[p].s:=t[p].s+int64(v)*(y-x+);
  if (l=x)and(r=y) then
  begin
   t[p].a:=t[p].a+v;
   exit;
  end;
  mid:=(l+r)>>;
  if y<=mid then update(l,mid,x,y,v,t[p].l)
   else if x>mid then update(mid+,r,x,y,v,t[p].r)
   else
   begin
    update(l,mid,x,mid,v,t[p].l);
    update(mid+,r,mid+,y,v,t[p].r);
   end;
 end;

 procedure query(l,r,x,y,p1,p2:longint);
 var mid:longint;
 begin
  if (l=x)and(r=y) then
  begin
   ans:=ans+t[p1].s-t[p2].s;
   exit;
  end;
  ans:=ans+(t[p1].a-t[p2].a)*(y-x+);
  mid:=(l+r)>>;
  if y<=mid then query(l,mid,x,y,t[p1].l,t[p2].l)
   else if x>mid then query(mid+,r,x,y,t[p1].r,t[p2].r)
   else
   begin
    query(l,mid,x,mid,t[p1].l,t[p2].l);
    query(mid+,r,mid+,y,t[p1].r,t[p2].r);
   end;
 end;

 procedure swap(var x,y:longint);
 var t:longint;
 begin
  t:=x; x:=y; y:=t;
 end;

 procedure qsort(l,r:longint);
 var i,j,mid:longint;
 begin
  i:=l; j:=r; mid:=d[(l+r)>>,];
  repeat
   while mid>d[i,] do inc(i);
   while mid<d[j,] do dec(j);
   if i<=j then
   begin
    swap(d[i,],d[j,]);
    swap(d[i,],d[j,]);
    swap(d[i,],d[j,]);
    swap(d[i,],d[j,]);
    inc(i); dec(j);
   end;
  until i>j;
  if l<j then qsort(l,j);
  if i<r then qsort(i,r);
 end;

 begin
  assign(input,'bzoj4826.in'); reset(input);
  assign(output,'bzoj4826.out'); rewrite(output);
  readln(n,que,p1,p2);
  for i:= to n do read(a[i]);
  top:=; stk[top]:=; a[]:=maxlongint;
  for i:= to n do
  begin
   while (top>)and(a[i]>a[stk[top]]) do dec(top);
   if top= then l[i]:=
    else l[i]:=stk[top];
   inc(top); stk[top]:=i;
  end;
  top:=; stk[top]:=n+; a[n+]:=maxlongint;
  for i:=n downto  do
  begin
   while (top>)and(a[i]>a[stk[top]]) do dec(top);
   if top= then r[i]:=n+
    else r[i]:=stk[top];
   inc(top); stk[top]:=i;
  end;
  for i:= to n do
  begin
   if (l[i]>)and(r[i]<=n) then
   begin
    inc(m); d[m,]:=l[i]; d[m,]:=r[i]; d[m,]:=r[i]; d[m,]:=p1;
   end;
   if l[i]> then
   begin
    inc(m); d[m,]:=l[i]; d[m,]:=i+; d[m,]:=r[i]-; d[m,]:=p2;
   end;
   if r[i]<=n then
   begin
    inc(m); d[m,]:=r[i]; d[m,]:=l[i]+; d[m,]:=i-; d[m,]:=p2;
   end;
  end;
  qsort(,m);
  j:=;
  for i:= to n do
  begin
   root[i]:=root[i-];
   while (j<=m)and(d[j,]=i) do
   begin
    update(,n,d[j,],d[j,],d[j,],root[i]);
    inc(j);
   end;
  end;
  for i:= to que do
  begin
   readln(x,y);
   ans:=(y-x)*p1;
   query(,n,x,y,root[y],root[x-]);
   writeln(ans);
  end;
  close(input);
  close(output);
 end.