POJ 3253 -- Fence Repair

Posted 2020-10-24 Amysear

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

 

题意：

有一个农夫要把一个木板钜成几块给定长度的小木板，每次锯都要收取一定费用，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：

使用贪心策略，跟Huffman树求最小码长的思想是一样的。

要使总费用最小，那么每次只选取最小长度的两块木板相加，再把这些“和”累加到总费用中即可

使用优先级队列Priority Queues维护一个小顶堆，priority_queue<int,vector<int>,greater<int> >

要注意写法“> >”，写成“>>”的话编译器会误认为是移位运算符

 

这样每次取两个队头元素（最小的两个int）

木板的块数N (1 ≤ N ≤ 20,000) ，每块木板的长度 L_i (1 ≤ L_i ≤ 50,000)

所以要使用 __int64进行钱数的加和计算，int会WA

 

优先级队列=》STL priority_queue

　　　　　　  STL -- heap结构及算法

 

 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         ///建立优先队列，优先级定义为数字小的优先级大
11         ///而且要注意写法“> >”，写成“>>”的话编译器会误认为是移位运算符
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();//取队头元素
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 }