Fence Repair

Fence Repair

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 32347		Accepted: 10391

Description

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.

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

C语言程序代码

/*
：：
		与NYOJ上的懒省事的小明那个题一样，只是说的不同而已。
题意：：
		大致为有一个农夫要把一个木板钜成几块给定长度的小木板，每次锯都要收取，
		定费用，这个费用就是当前锯的这个木版的长度给定各个要求的小木板的长度，
		及小木板的个数n，求最小费用。
解题思路：：
		貌似这个题很绕口，理解起来比较难，但学长给出另一种思路，即倒着推，就是，将
		木板占合起来，所花费的最小费用。
		1、先粘合最短的两块，因为，短的花费的少。
		2、用队列将其从小到大排序，
		3、每次去前两块，将其加和后再输入，直到粘为一块，
		4、输出和即可。 


*/
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<queue>
using namespace std;
int a[51000];
priority_queue<long long int,vector<long long int>,greater<long long int> >q;
int main(){
	int t,n,m,i,j,k;
	long long int sum,d,b,c;
	while(scanf("%d",&t)!=EOF)
	{
		sum=d=b=c=0;
		for(i=0;i<t;i++)
		{
			scanf("%d",&a[i]);
			q.push(a[i]);
		}
		while(q.size()!=1)
		{
			d=q.top() ;
			q.pop() ;
			b=q.top();
			q.pop() ;
			c=d+b;
			sum+=c;
			q.push(c);
		}
		printf("%lld\n",sum);
	}
	return 0;
}