【CF675E】Trains and Statistic（贪心，DP，线段树优化）

题意：
a[i]表示从第i个车站可以一张票到第[i+1,a[i]]这些车站;
p[i][j]表示从第i个车站到第j个车站的最少的票数,现在要求∑dp[i][j](1<=i<=n,i<j<=n);

思路：从I开始走，在i+1到a[i]之间一定会到使a[j]最大的j，因为要使步数最小，接下来能走得更快
     区间询问最值用RMQ与线段树都可以
     dp[i]表示dp[i,i+1],dp[i,i+2]...dp[i,n]这些值的和
        dp[i]=dp[k]+(n-i)-(a[i]-k),k为[i+1,a[i]]中使a[k]最大的k
        n-i：有dp[i,i+1],dp[i,i+2]...dp[i,n]一共n-i个状态要从i走到k
        -(a[i]-k)：dp[i,k+1]到 dp[i,a[i]]长度与之前相比没有变化

 type ok=record
          s:int64;
          a:longint;
         end;

 const oo=;

 var dp:array[..]of int64;
     a:array[..]of longint;
     tree:array[..]of ok;
     n,i,k:longint;
     ans:int64;

 procedure pushup(p:longint);
 begin
  if tree[p<<].s>tree[p].s then tree[p]:=tree[p<<];
  if tree[p<<+].s>tree[p].s then tree[p]:=tree[p<<+];
 end;

 procedure build(l,r,p:longint);
 var mid:longint;
 begin
  tree[p].s:=a[l];
  tree[p].a:=l;
  if l=r then exit;
  mid:=(l+r)>>;
  build(l,mid,p<<);
  build(mid+,r,p<<+);
  pushup(p);
 end;

 function query(l,r,x,y,p:longint):ok;
 var mid:longint;
     t:int64;
     q,tmp:ok;

 begin
  if (l=x)and(r=y) then exit(tree[p]);
  mid:=(l+r)>>;
  t:=;
  if (x>=l)and(y<=mid) then
  begin
   q:=query(l,mid,x,y,p<<);
   if q.s>t then begin t:=q.s; tmp:=q; end;
  end else
  if (x>mid)and(y<=r) then
  begin
   q:=query(mid+,r,x,y,p<<+);
   if q.s>t then begin t:=q.s; tmp:=q; end;
  end else
  begin
   q:=query(l,mid,x,mid,p<<);
   if q.s>t then begin t:=q.s; tmp:=q; end;
   q:=query(mid+,r,mid+,y,p<<+);
   if q.s>t then begin t:=q.s; tmp:=q; end;
  end;
  exit(tmp);
 end;

 begin
  //assign(input,'1.in'); reset(input);
  //assign(output,'1.out'); rewrite(output);
  readln(n);
  for i:= to n- do read(a[i]);
  a[n]:=n;
  build(,n,);

  for i:=n- downto  do
  begin
   k:=query(,n,i+,a[i],).a;
   dp[i]:=dp[k]+(n-i)-(a[i]-k);
  end;
  for i:= to n do ans:=ans+dp[i];
  writeln(ans);
  //close(input);
  //close(output);
 end.