Binary heap (1): Basic

It turns out that although I know more about the heap theory, I only sorted several heap jobs. I 'd like to review it recently.

Poj3253

{FJ needs to cut a wooden rod into N wooden bars of a given length. Each time the cost is the length of the wooden rod. The total cost and the merged fruit are basically the same, but the title description is inverse, you need to think about cutting as merging. You can combine the smallest wooden bars and record the most cost. Although queuing is the best, it is used to practice basic heap operations. It supports filtering, returns and deletes the minimum element} var A: array [0 .. 1000000] of longint; heapsize: longint; m, n, min1, min2: longint; I, J, K: longint; ans: int64; Procedure heap_down (K: longint ); vaR I, X: longint; begin X: = A [k]; I: = K * 2; While I <= heapsize Do Begin If (I  If A [I] <X then begin a [k]: = A [I]; K: = I; I: = I * 2; End Else   Break ; End; A [k]: = x; end; Procedure swap (var a, B: longint); var C: longint; begin C: = A; A: = B; b: = C; end; function get_min: longint; var X: longint; begin swap (A [1], a [heapsize]); Dec (heapsize ); heap_down (1); exit (A [heapsize + 1]); end; procedure insert (key: longint); var I: longint; begin Inc (heapsize); I: = heapsize; While A [I Div 2]> key Do Begin a [I]: = A [I Div 2]; I: = I Div 2; end; A [I]: = key; end; begin readln (N ); For I: = 1 to n Do Read (A [I]); heapsize: = N; For I: = heapsize Div 2 downto 1 Do Heap_down (I ); For I: = 1 to n-1 Do Begin min1: = get_min; min2: = get_min; Inc (ANS, min1 + min2); insert (min1 + min2); end; writeln (ANS); end.

Poj2051 (wa'sCode)

{Question: Enter n task numbers and latencies. After each task is executed, the latency increases at the start, output the serial number of the previous m tasks in sequence (in the order of serial numbers when the latency is the same ). Create a small root heap. Each time you output the heap top element and add the value of the heap top element keyword and filter it down, I don't know why WA2011-06-19} type data = record delay, time, num: longint end; var: array [0 .. 100000] of data; I, J, K, M, N: longint; s: String [7]; C: Char ; Procedure heap_down (K: longint); var X: data; I: longint; begin X: = A [k]; I: = K * 2; While I <= N Do Begin If (I <n) and (A [I]. delay> A [I + 1]. delay) or (A [I]. delay = A [I + 1]. delay) and (A [I]. num> A [I + 1]. num) Then Inc (I ); If X. Delay> A [I]. Delay then begin a [k]: = A [I]; K: = I; I: = I * 2; End Else  Break ; End; A [k]: = x; end; Procedure inc_key (K, add: longint); begin Inc (A [K]. delay, add); heap_down (1); end; begin While Not EOF Do Begin N: = 0; fillchar (, Sizeof (A), 0); read (c ); While C = 'R'   Do Begin read (C, C); Inc (n); readln (A [n]. num, A [n]. time); A [n]. delay: = A [n]. time; read (c); end; readln (m ); For I: = N Div 2 downto 1 Do Heap_down (I ); For I: = 1 to m Do Begin writeln (A [1]. Num); inc_key (1, a [1]. Time); end.