poj3709

首先我们发现将一段数变为同一个数比间隔着搞肯定优，
因为数列是升序的，
然后不难得到方程式f[i]=min(f[j]+sum[i]-sum[j]-(i-j)*a[j+1]) (i-j>=m)
简单的斜率优化不多说
注意这道题最优解有选择范围，也就是说要延迟入队

var a,s,f:array[0..500010] of int64;
    q:array[0..500010] of longint;
    j,t,i,n,m,h,r:longint;

function G(j,k:int64):int64;
  begin
    exit(f[j]-f[k]+s[k]-s[j]+a[j+1]*j-a[k+1]*k);
  end;

function P(j,k:int64):int64;
  begin
    exit(a[j+1]-a[k+1]);
  end;

function value(i,j:int64):int64;
  begin
    exit(f[j]+s[i]-s[j]-a[j+1]*(i-j));
  end;

begin
  readln(t);
  while t>0 do
  begin
    dec(t);
    readln(n,m);
    s[0]:=0;
    for i:=1 to n do
    begin
      read(a[i]);
      s[i]:=s[i-1]+a[i];
    end;
    f[0]:=0;
    q[0]:=0;
    h:=0;
    r:=0;
    for i:=1 to n do
    begin
      while (h<r) and (G(q[h+1],q[h])<=int64(i)*P(q[h+1],q[h])) do inc(h);
      f[i]:=value(i,q[h]);
      if (i>=2*m-1) and (i<>n) then
      begin
        while (h<r) and (G(i-m+1,q[r])*P(q[r],q[r-1])<=G(q[r],q[r-1])*P(i-m+1,q[r])) do dec(r);
        inc(r);
        q[r]:=i-m+1;
      end;
    end;
    writeln(f[n]);
  end;
end.