Fence Repair

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 27055		Accepted: 8800

Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needs N (1 ≤ N ≤ 20,000) planks of wood, each having some integer length L_i (1 ≤ L_i ≤
50,000) units. He then purchases a single long board just long enough to saw into the N planks (i.e., whose length is the sum of the lengths L_i). FJ is ignoring the "kerf", the extra length lost to sawdust when a sawcut is made;
you should 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 of the N-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 create the N planks. FJ knows that he can cut the board in various different orders which will
result in different charges since the resulting intermediate planks are of different lengths.

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

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

3

8

5

8

34

哈曼夫树。

题意 ：把一根长度为x的木板切成n段。每段的长度已给出来。切某块木板时。花费的费用为该木板的长度。一開始以为仅仅要每次选出最长的木板然后从总的中切掉就是最小花费。实际上不是，其有用脑子想想也知道了，假设总长非常长的话，而须要的木板的长度又非常小，每次从总长中切掉花费将会非常大，所以这样的思路是不正确的，看了讨论区才明确，我们能够把切这个过程逆回去。就是把这n段木板连接起来，每次花费的钱为待连接的两段木板的长度。（细致想想，切跟连花费的钱是同样的，即是等价的，画一下图能够看出来） 然后有一种思想是：初始化一个最小堆，每次从堆中选出最小的两段木板进行连接。连接完之后在放入到堆中。总共n-1次连接就可以

#include <cstdio>

#include <iostream>

#include <algorithm>

#include <cstring>

#include <cctype>

#include <cmath>

#include <cstdlib>

#include <vector>

#include <queue>

#include <set>

#include <map>

#include <list>

#define ll long long

using namespace std;

const int INF = 0x3f3f3f3f;

ll a[20010];

int main()

{

	int n;

	while(~scanf("%d",&n))

	{

		for(int i=0;i<n;i++)

			scanf("%lld",a+i);

		make_heap(a,a+n,greater<int>());

		int t=n-1;ll ans=0,tem;

		while(t--)

		{

			tem=a[0];pop_heap(a,a+n,greater<int>());--n;

			tem+=a[0];pop_heap(a,a+n,greater<int>());--n;

			ans+=tem;a[n++]=tem;

			push_heap(a,a+n,greater<int>());

		}

		printf("%lld\n",ans);

	}

	return 0;

}