POJ 3253 贪心

poj.org/problem?id=3253

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 Li (1 ≤ Li ≤ 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 Li). 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.

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

3
8
5
8

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 should be used to cut the board into pieces measuring 13 and 8. The second cut will cost 13, and should be used to cut the 13 into 8 and 5. This would cost 21+13=34. If the 21 was cut into 16 and 5 instead, the second cut would cost 16 for a total of 37 (which is more than 34).

题目大意：给你n块木板的长度，设其总长度为L，问你把L经过n-1次划分划分为n块木板的最小代价。划分长度为L的木板的代价为L。

思路：霍夫曼编码问题。和石子归并是一个道理，正着贪心的话自然想到每次划分都分出来一块当前最大的木板，反过来呢，就是把木板合并起来，每次从所有木板中选取两块木板，删除这两块木板，再将合并的木板加入到模板序列中，他们长度之和是这次合并的代价。可以用优先队列实现。注意会爆int。

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#define INF 0x3f3f3f3f
typedef long long ll;
using namespace std;

int main()
{
	priority_queue<ll,vector<ll>,greater<ll> >q;
	int n;
	scanf("%d",&n);
	ll temp;
	for(int i=0;i<n;i++)
	{
		scanf("%lld",&temp);
		q.push(temp);
	}
	ll sum=0;
	while(!q.empty())
	{
		ll t=q.top();
		q.pop();
		if(q.empty())
			break;
		t+=q.top();
		q.pop();
		sum+=t;
		q.push(t);
	}
	printf("%lld\n",sum);
	return 0;
}