bzoj2006

论将区间和转化为前缀和的重要性
这题一旦转化为前缀和就非常明了了
一段区间[l,r]的美妙程度就等于s[r]-s[l-1]
对于这种无法计算出所有方案而取前k大的题目，我们一般分类别然后利用类别内的单调性用堆维护
对于以i开头的区间，显然最美的长度在[l,r]之间的区间和
显然是max(s[i+l-1~i+r-1])，设这个最美的为k
次美的就是max(max(s[i+l-1~k-1]),max(s[k+1~i+r-1]),以此类推
不难想到用ST预处理区间前缀最大值
然后用大根堆维护开头为i,末端点位置在[p,q]上的最大值，
每次取出当前堆最大值，设在k取到最大值，就将这个区间裂位[p,k-1],[k+1,q]两个区间加入堆
一共进行k次，总的复杂度大约是O(klogn+nlogn)
讲的可能不是很清楚，具体见程序吧

 type node=record
        tl,tr,st,loc:longint;
        num:int64;
      end;

 var heap:array[..] of node;
     d:array[..] of longint;
     f:array[..,..] of longint;
     s:array[..] of longint;
     n,k,l,r,i,j,x,y,t,p:longint;
     ans:int64;

 function max(a,b:longint):longint;
   begin
     if s[a]>s[b] then exit(a) else exit(b);
   end;

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

 function ask(x,y:longint):longint;
   var k:longint;
   begin
     k:=trunc(ln(y-x+)/ln());
     exit(max(f[x,k],f[y-d[k]+,k]));
   end;

 procedure insert(i,x,y:longint);
   begin
     heap[t].tl:=x;
     heap[t].tr:=y;
     heap[t].st:=i;
     heap[t].loc:=ask(x,y);
     heap[t].num:=s[heap[t].loc]-s[i-];
   end;

 procedure swap(var a,b:node);
   var c:node;
   begin
     c:=a;
     a:=b;
     b:=c;
   end;

 procedure up(i:longint);
   var j:longint;
   begin
     j:=i shr ;
     while j> do
     begin
       if heap[i].num>heap[j].num then
       begin
         swap(heap[i],heap[j]);
         i:=j;
         j:=i shr ;
       end
       else break;
     end;
   end;

 procedure sift(i:longint);
   var j:longint;
   begin
     j:=i shl ;
     while j<=t do
     begin
       if (j<t) and (heap[j].num<heap[j+].num) then inc(j);
       if heap[i].num<heap[j].num then
       begin
         swap(heap[i],heap[j]);
         i:=j;
         j:=i shl ;
       end
       else break;
     end;
   end;

 begin
   readln(n,k,l,r);
   for i:= to n do
   begin
     readln(x);
     s[i]:=s[i-]+x;
     f[i,]:=i;
   end;
   t:=trunc(ln(n)/ln());
   d[]:=;
   for i:= to t do
     d[i]:=d[i-]*;
   for j:= to t do
     for i:= to n do
       if i+d[j]-<=n then
         f[i,j]:=max(f[i,j-],f[i+d[j-],j-])
       else break;
   t:=;
   for i:= to n-l+ do
   begin
     inc(t);
     insert(i,i+l-,min(i+r-,n));
     up(i);
   end;
   for i:= to k do
   begin
     p:=heap[].loc;
     x:=heap[].tl;
     y:=heap[].tr;
     j:=heap[].st;
     ans:=ans+int64(heap[].num);
     swap(heap[],heap[t]);
     dec(t);
     sift();
     if p>x then
     begin
       inc(t);
       insert(j,x,p-);
       up(t);
     end;
     if p<y then
     begin
       inc(t);
       insert(j,p+,y);
       up(t);
     end;
   end;
   writeln(ans);
 end.