pku3418 Double Queue
需要这样一个数据结构，支持如下操作
1：插入优先级为p，编号为k的节点
2：查询优先级最高的节点，输出编号并删除
3：查询优先级最低的节点，输出编号并删除
用一颗SBT即可完美解决问题，没什么好说的，多说无益~~~
View Code
1 program pku3481(input,output); 2 var 3 left,right,key,s,th:array[0..200000] of longint; 4 tot,root:longint; 5 procedure left_rotate(var t:longint); 6 var 7 k:longint; 8 begin 9 k:=right[t]; 10 right[t]:=left[k]; 11 left[k]:=t; 12 s[k]:=s[t]; 13 s[t]:=s[left[t]]+s[right[t]]+1; 14 t:=k; 15 end;{ left_rotate } 16 procedure right_rotate(var t:longint); 17 var 18 k:longint; 19 begin 20 k:=left[t]; 21 left[t]:=right[k]; 22 right[k]:=t; 23 s[k]:=s[t]; 24 s[t]:=s[left[t]]+s[right[t]]+1; 25 t:=k; 26 end;{ right_rotate } 27 procedure maintain(var t:longint;flag:boolean); 28 begin 29 if not flag then 30 begin 31 if s[left[left[t]]]>s[right[t]] then 32 right_rotate(t) 33 else 34 if s[right[left[t]]]>s[right[t]] then 35 begin 36 left_rotate(left[t]); 37 right_rotate(t); 38 end 39 else 40 exit; 41 end 42 else 43 begin 44 if s[right[right[t]]]>s[left[t]] then 45 left_rotate(t) 46 else 47 if s[left[right[t]]]>s[left[t]] then 48 begin 49 right_rotate(right[t]); 50 left_rotate(t); 51 end 52 else 53 exit; 54 end; 55 maintain(left[t],false); 56 maintain(right[t],true); 57 maintain(t,false); 58 maintain(t,true); 59 end;{ maintain } 60 procedure insert(var now,k,u:longint); 61 begin 62 if now=0 then 63 begin 64 inc(tot); 65 now:=tot; 66 s[now]:=1; 67 left[now]:=0; 68 right[now]:=0; 69 key[now]:=k; 70 th[now]:=u; 71 end 72 else 73 begin 74 inc(s[now]); 75 if k<key[now] then 76 insert(left[now],k,u) 77 else 78 insert(right[now],k,u); 79 maintain(now,k>=key[now]); 80 end; 81 end;{ insert } 82 function delete(var t:longint;k:longint):longint; 83 begin 84 dec(s[t]); 85 if (k=key[t])or((k<key[t])and(left[t]=0))or((k>key[t])and(right[t]=0)) then 86 begin 87 delete:=key[t]; 88 if left[t]*right[t]=0 then 89 t:=left[t]+right[t] 90 else 91 key[t]:=delete(left[t],k+1); 92 end 93 else 94 if k<key[t] then 95 delete:=delete(left[t],k) 96 else 97 delete:=delete(right[t],k); 98 end;{ delete } 99 function get_max_number(t:longint):longint; 100 begin 101 if t=0 then 102 exit(19950714); 103 get_max_number:=get_max_number(right[t]); 104 if get_max_number=19950714 then 105 get_max_number:=t; 106 end;{ get_max_number } 107 function get_min_number(t:longint):longint; 108 begin 109 if t=0 then 110 exit(19950714); 111 get_min_number:=get_min_number(left[t]); 112 if get_min_number=19950714 then 113 get_min_number:=t; 114 end;{ get_min_number } 115 procedure main; 116 var 117 x,y,z:longint; 118 begin 119 root:=0; 120 tot:=0; 121 read(x); 122 while x<>0 do 123 begin 124 if x=1 then 125 begin 126 readln(y,z); 127 insert(root,z,y); 128 end; 129 if x=2 then 130 if s[root]=0 then 131 writeln(0) 132 else 133 begin 134 y:=get_max_number(root); 135 writeln(th[y]); 136 delete(root,key[y]); 137 end; 138 if x=3 then 139 if s[root]=0 then 140 writeln(0) 141 else 142 begin 143 y:=get_min_number(root); 144 writeln(th[y]); 145 delete(root,key[y]); 146 end; 147 read(x); 148 end; 149 end;{ main } 150 begin 151 main; 152 end.
