POJ 3253 Fence Repair (Analog Huffman tree + priority queue)

Test instructions: If you want to cut a wood bar of length A to cost a, ask for the minimum cost of cutting out the required n bars.

Idea: Simulate the Huffman tree, select the minimum two numbers at a time to add the results, and then add the two numbers and join the queue.

Note the use of priority_queue, prototypes:

1 priority_queue<type> q;   2 priority_queue<type,deque<type>,comp> q;  

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Where type is the type, comp is the comparison struct, and the comparison function is its bracket overload, such as the comp structure of the type int from small to large:

1 struct Comp   2 {  3     booloperator() (const ll& A,const Const  4     {  5         return a>b;   6     }  7 };  

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It is also important to note that there is no direct while (!q.empty ()), because each time you select two numbers, if the queue has only one number to take two numbers, then the program will explode, should use a while (1) loop, when the number is taken out immediately determine if (Q.empty ()) Break

Cannot directly simulate Huffman tree, direct Huffman will time out.

1 #pragmaComment (linker, "/stack:1024000000,1024000000")2#include <iostream>3#include <cstdio>4#include <cstring>5#include <cmath>6#include <math.h>7#include <algorithm>8#include <queue>9#include <Set>Ten#include <bitset> One#include <map> A#include <vector> -#include <stdlib.h> -#include <stack> the using namespacestd; - #definePI ACOs (-1.0) - #defineMax (a) (a) > (b)? (a): (b) - #defineMin (a) (a) < (b)? (a): (b) + #definell Long Long - #defineEPS 1e-10 + #defineMOD 1000000007 A #defineN 1000000 at #defineINF 1e12 - ll N; - structcomp{ -    BOOL operator()(Constll &a,Constll &b)Const{ -       returnA>b; -    } in }; - intMain () to { +     while(SCANF ("%i64d", &n) = =1){ -Priority_queue<ll,deque<ll>,comp>Q; the        for(intI=1; i<=n;i++){ * ll X; $scanf"%i64d",&x);Panax Notoginseng Q.push (x); -       } thell ans=0; +        while(1){ All l=q.top (); Q.pop (); the          if(Q.empty ()) Break; +ll r=q.top (); Q.pop (); -ans+=L; $ans+=R; $Q.push (L +R); -       } -printf"%i64d\n", ans); the    } -     return 0;Wuyi}

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POJ 3253 Fence Repair (Analog Huffman tree + priority queue)