最大连续长度不超过m的子序列和

题目大意

给定长度为n的序列,取出其中长度不超过m的连续一段,求最大

题解

ans=max{sum[i]−sum[j−1]}   (i>=j−1且1<=j−i+1<=m)
很明显这是 O(N2) 的暴力
对于位置i它前面sum[j-1]最小即可,也就是静态查询区间最小(ST表/线段树)
我们优化到 O(NlogN)
对于一个前缀和,它的下标小于后面的下标,并且它的权值也小于后面的权值,即 i<j且sum[i]<sum[j] 那么我们舍去i,单调队列维护
我们优化到 O(N)

const
 maxn=200005;
var
 x,sum:array[0..maxn]of longint;
 w:array[0..4*maxn,1..3]of longint;
 i,j,k:longint;
 n,m,ans:longint;
function max(a,b:longint):longint;
begin
 if a>b then exit(a) else exit(b);
end;

function min(a,b:longint):longint;
begin
 if athen exit(a) else exit(b);
end;

procedure build(a,l,r:longint);
var mid:longint;
begin
 w[a,1]:=l; w[a,2]:=r; w[a,3]:=0;
 if l=r then begin w[a,3]:=sum[l]; exit; end;
 mid:=(l+r)>>1;
 build(a<<1,l,mid); build(a<<1+1,mid+1,r);
 w[a,3]:=min(w[a<<1,3],w[a<<1+1,3]);
end;

function query(a,l,r:longint):longint;
var mid:longint;
begin
 if (w[a,1]=l)and(w[a,2]=r) then exit(w[a,3]);
 mid:=(w[a,1]+w[a,2])>>1;
 if r<=mid then exit(query(a<<1,l,r)) else
 if l>mid then exit(query(a<<1+1,l,r))
 else exit(min(query(a<<1,l,mid),query(a<<1+1,mid+1,r)));
end;

begin
 readln(n,m);
 for i:=1 to n do
  begin
   read(x[i]); sum[i]:=sum[i-1]+x[i];
  end;
 build(1,1,n);
 ans:=0;
 for i:=n downto 2 do
  begin
   if i-m>0
   then ans:=max(sum[i]-query(1,i-m,i-1),ans)
   else ans:=max(sum[i]-query(1,1,i-1),ans);
  end;
 ans:=max(ans,x[1]);
 writeln(ans);
end.