poj3253 Fence Repair STL優先隊列

標籤：stl   優先隊列   

轉載請註明出處：http://blog.csdn.net/u012860063

題目連結：http://poj.org/problem?id=3253

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 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

3858

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

大致題意：

有一個農夫要把一個木板钜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

 

提示：

以

3

5 8 5為例：

先從無限長的木板上鋸下長度為 21 的木板，花費 21

再從長度為21的木板上鋸下長度為5的木板，花費5

再從長度為16的木板上鋸下長度為8的木板，花費8

總花費 = 21+5+8 =34

 

解題思路：

利用Huffman思想，要使總費用最小，那麼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

本題雖然利用了Huffman思想，但是直接用HuffmanTree做會逾時，可以用優先隊列做

代碼如下：

/*STL 優先隊列*/#include <cstdio>#include <queue>#include <vector>#include <iostream>using namespace std;int main(){int n;//需要切割的木板個數__int64 temp,a,b,mincost;while(~scanf("%d",&n)){//定義從小到大的優先隊列，可將greater改為less，即為從大到小priority_queue<int, vector<int>, greater<int> > Q;while(!Q.empty())//清空隊列Q.pop();for(int i = 1; i <= n; i++){scanf("%I64d",&temp);Q.push(temp);//輸入要求的木板長度（費用）併入隊}mincost = 0;//最小費用初始為零while(Q.size() > 1)//當隊列中小於等於一個元素時跳出{a = Q.top();//得到隊首元素的值，並使其出隊Q.pop();b = Q.top();//兩次取隊首，即得到最小的兩個值Q.pop();Q.push(a+b);//把兩個最小元素的和入隊mincost +=a+b;}printf("%I64d\n",mincost);}return 0;}