poj_3253 Fence Repair

Fence Repair

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

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).

 

题意：要切N块板子，将长度为L的板子切成两块花费L元，给出N块板子的长度，求最小花费

 

解析：Huffman tree 思想，每次将最短的两块板子取出，再将两者长度之和累加到费用中，再将两者长度之和作为新一块板子的长度并把这块板子放入剩余板子的集合中，重复以上操作直到只剩一块板子。

具体操作可以利用STL大法，强行优先队列

代码如下：

#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<queue>
using namespace std;
int n;
long long ans;
priority_queue<long long ,vector<long long>,greater<long long> > Q;
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
long long a;
scanf("%lld",&a);
Q.push(a);
}
while(Q.size()!=1)
{
long long u=Q.top();Q.pop();
long long v=Q.top();Q.pop();
ans+=(u+v);
u+=v;
Q.push(u);
}
printf("%lld",ans);
return 0;
}