使用可持久化+trie解决特定问题的高效算法-优快云博客

本文详细介绍了如何通过结合可持久化数据结构和Trie树来解决特定类型的问题，包括实现步骤、关键算法和优化策略。通过实例演示了这种算法的应用，展示了其实现的高效性和灵活性。

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这题做法很多
可以通过类似noi超级钢琴那道题目的做法用可持久化+trie来做
还可以直接在trie树上维护size域然后类似查找k大的做法做
总之还是比较水的

  1 type node=record
  2        kth,num,ans:longint;
  3      end;
  4 
  5 var heap:array[0..100010] of node;
  6     son:array[0..3000010,0..1] of longint;
  7     size:array[0..3000010] of longint;
  8     a:array[0..100010] of longint;
  9     t,i,n,m:longint;
 10 
 11 procedure swap(var a,b:node);
 12   var c:node;
 13   begin
 14     c:=a;
 15     a:=b;
 16     b:=c;
 17   end;
 18 
 19 procedure add(x:longint);
 20   var i,p,y:longint;
 21   begin
 22     p:=1;
 23     for i:=30 downto 0 do
 24     begin
 25       y:=x and (1 shl i);
 26       if y>1 then y:=1;
 27       if son[p,y]=0 then
 28       begin
 29         inc(t);
 30         son[p,y]:=t;
 31       end;
 32       p:=son[p,y];
 33       inc(size[p]);
 34     end;
 35   end;
 36 
 37 function ask(x,k:longint):longint;
 38   var p,i,y:longint;
 39   begin
 40     p:=1;
 41     ask:=0;
 42     for i:=30 downto 0 do
 43     begin
 44       y:=x and (1 shl i);
 45       if y>1 then y:=1;
 46       if size[son[p,y]]>=k then p:=son[p,y]
 47       else begin
 48         ask:=ask+1 shl i;
 49         k:=k-size[son[p,y]];
 50         p:=son[p,1-y];
 51       end;
 52     end;
 53   end;
 54 
 55 procedure sift(i:longint);
 56   var j:longint;
 57   begin
 58     j:=i shl 1;
 59     while j<=t do
 60     begin
 61       if (j+1<=t) and (heap[j].ans>heap[j+1].ans) then inc(j);
 62       if heap[i].ans>heap[j].ans then
 63       begin
 64         swap(heap[i],heap[j]);
 65         i:=j;
 66         j:=i shl 1;
 67       end
 68       else break;
 69     end;
 70   end;
 71 
 72 begin
 73   readln(n,m);
 74   t:=1;
 75   for i:=1 to n do
 76   begin
 77     readln(a[i]);
 78     add(a[i]);
 79   end;
 80   for i:=1 to n do
 81   begin
 82     heap[i].num:=a[i];
 83     heap[i].kth:=2;
 84     heap[i].ans:=ask(a[i],2);
 85   end;
 86   t:=n;
 87   for i:=n div 2 downto 1 do
 88     sift(i);
 89   for i:=1 to 2*m do
 90   begin
 91     if i mod 2=1 then write(heap[1].ans,' ');  //注意会被重复计算
 92     if heap[1].kth=n then
 93     begin
 94       swap(heap[1],heap[t]);
 95       dec(t);
 96     end
 97     else begin
 98       inc(heap[1].kth);
 99       heap[1].ans:=ask(heap[1].num,heap[1].kth);
100     end;
101     sift(1);
102   end;
103 end.