Poj 3253 fence repair is similar to the greedy idea of the Harman tree.

Fence repair

Time limit:2000 ms		Memory limit:65536 K
Total submissions:24550		Accepted:7878

Description

Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needsN(1 ≤N≤ 20,000) planks of wood, each having some integer LengthLi(1 ≤Li≤ 50,000) units. He then purchases a single long board just long enough to saw intoNPlanks (I. e., whose length is the sum of the lengthsLi). Fj is ignoring the "kerf", the extra length lost to sawdust when a sawcut is made; You shoshould ignore it, too.

FJ sadly realizes that he doesn't own a saw with which to cut the wood, so he mosies over to farmer don's farm with this long board and politely asks if he may borrow a saw.

Farmer Don, a closet capitalist, doesn't lend FJ a saw but instead offers to charge Farmer John for each ofN-1 cuts in the plank. The charge to cut a piece of wood is exactly equal to its length. Cutting a plank of length 21 costs 21 cents.

Farmer Don then lets farmer john decide the order and locations to cut the plank. Help Farmer John determine the minimum amount of money he can spend to createNPlanks. fj knows that he can cut the board in varous different orders which will result in different charges since the resulting intermediate planks are of different lengths.

Input

Line 1: One integer N, The number of planks
Lines 2 .. N+ 1: each line contains a single integer describing the length of a needed plank

Output

Line 1: One INTEGER: the minimum amount of money he must spend to make N-1 cuts

Sample Input

3858

Sample output

34

Hint

He wants to cut a board of length 21 into pieces of lengths 8, 5, and 8.
The original Board measures 8 + 5 + 8 = 21. the first cut will cost 21, and shoshould be used to cut the Board into pieces measuring 13 and 8. the second cut will cost 13, and shoshould be used to cut the 13 into 8 and 5. this wocould cost 21 + 13 = 34. if the 21 was cut into 16 and 5 instead, the second cut wocould cost 16 for a total of 37 (which is more than 34 ).

Source

Usaco 2006 November gold
Question

Think about it using the idea similar to the Harman tree. Select two smaller vertices to merge each time, and then return the merged vertex to the original set. Use a small root heap for maintenance.

Note: first, the priority queue of C ++ is a large top heap by default, so it needs to be rewritten. Second, the modified heap does not have the clear () method, so remember to clear the queue!

Sample Code

/** =================================================== ================================================ # Copyright Notice # copyright (c) 2014 All Rights Reserved # ---- stay hungry Stay Foolish ---- ##@ Author: Shen # @ name: poj 3253 # @ file: G: \ My source code \ [ACM] training \ 0624-basics \ poj3253.cpp # @ Date: # @ algorithm: greedy = ========================================================= **/ // # pragma GCC OPT Imize ("O2") // # pragma comment (linker, "/Stack: 1024000000,1024000000 ") # include <queue> # include <cstdio> # include <string> # include <cstring> # include <iomanip> # include <iostream> # include <algorithm> using namespace STD; template <class T> inline bool updatemin (T & A, t B) {return A> B? A = B, 1: 0;} template <class T> inline bool updatemax (T & A, t B) {return a <B? A = B, 1: 0;} typedef long int64; int n, l; void solve () {priority_queue <int, vector <int>, greater <int> PQ; for (INT I = 0; I <n; I ++) {scanf ("% d", & L); PQ. push (l);} int64 ans = 0; If (n = 1) ans = L; while (PQ. size ()> 1) {int A = PQ. top (); PQ. pop (); int B = PQ. top (); PQ. pop (); int c = a + B; ans + = C; PQ. push (c);} printf ("% d \ n", ANS);} int main () {While (~ Scanf ("% d", & N) solve (); Return 0 ;}