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[..] of int64;
     q:array[..] 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+]*j-a[k+]*k);
   end;

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

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

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