ted eight out wan sum some cto 建立 hint

Fence Repair

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 55661		Accepted: 18331

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).

Source

USACO 2006 November Gold

題意：

有一個農夫要把一個木板鉅成幾塊給定長度的小木板，每次鋸都要收取一定費用，這個費用就是當前鋸的這個木版的長度

給定各個要求的小木板的長度，及小木板的個數n，求最小費用

解題思路：

使用貪心策略，跟Huffman樹求最小碼長的思想是一樣的。

要使總費用最小，那麽每次只選取最小長度的兩塊木板相加，再把這些“和”累加到總費用中即可

使用優先級隊列Priority Queues維護一個小頂堆，priority_queue<int,vector<int>,greater<int> >

要註意寫法“> >”，寫成“>>”的話編譯器會誤認為是移位運算符

這樣每次取兩個隊頭元素（最小的兩個int）

木板的塊數N (1 ≤ N ≤ 20,000) ，每塊木板的長度 L_i (1 ≤ L_i ≤ 50,000)

所以要使用 __int64進行錢數的加和計算，int會WA

優先級隊列=》STL priority_queue

　　　　　　 STL -- heap結構及算法

 1 #include<queue>
 2 #include<iostream>
 3 using namespace std;
 4 int main()
 5 {
 6     int n;
 7     while(cin>>n)
 8     {
 9         __int64 sum = 0;
10         ///建立優先隊列，優先級定義為數字小的優先級大
11         ///而且要註意寫法“> >”，寫成“>>”的話編譯器會誤認為是移位運算符
12         priority_queue<int,vector<int>,greater<int> > planks;
13         for(int i=0;i<n;i++)
14         {
15             int temp;
16             cin>>temp;
17             planks.push(temp);
18         }
19         while(planks.size() != 1)
20         {
21             int temp = 0;
22             temp += planks.top();//取隊頭元素
23             planks.pop();
24             temp += planks.top();
25             planks.pop();
26             sum+=temp;
27             planks.push(temp);
28         }
29 
30         cout<<sum<<endl;
31 
32 
33     }
34     return 0;
35 }

POJ 3253 -- Fence Repair