20140822 【 优先队列 】 POJ 3253 Fence Repair

Online JudgeProblem SetAuthorsOnline ContestsUserWeb Board
Home Page
F.A.Qs
Statistical Charts

Problems
Submit Problem
Online Status
Prob.ID:

Register
Update your info
Authors ranklist

Current Contest
Past Contests
Scheduled Contests
Award Contestwilson1068      Log Out
Mail:0(0)
Login Log      Archive

Language:Default

Fence Repair

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 26478 Accepted: 8589

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

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

[Submit]   [Go Back]   [Status]   [Discuss]

Home Page   Go Back  To top

---

All Rights Reserved 2003-2013 Ying Fuchen,Xu Pengcheng,Xie Di
Any problem, Please Contact Administrator

写这篇主要是为了记录 优先队列 的用法：

priority_queue<int, vector<int>, greater<int> > a;

第二参数为容器类型；

第三参数为比较函数；

建立出来的是升序的序列，【top】为首部最小值；

priority_queue<int>  b;

建立出来的是降序的序列，【top】为首部最大值；

/**

注意：

1.升序优先队列，一定要注意隔开最后的 "> >" ；

否则编译器会认为这是 位运算的右移“>>”，或者 cin的重载的读入操作符“>>”。

2.优先队列没有迭代器，想要遍历优先队列，就要不停地 top -> pop -> top -> pop ......

注意选择 恢复。

*/

#include <stdio.h>#include <string.h>#include <algorithm>#include <iostream>#include <queue>#include <vector>using namespace std;priority_queue<int, vector<int>, greater<int> > a;int main(){       long long n, x, y, ans;    while( EOF != scanf("%lld", &n) ){        for(int i=0; i<n; i++){            scanf("%lld", &x);            a.push(x);        }        ans = x = 0;        while( a.size()>1 ){            x = a.top();    a.pop();            x += a.top();   a.pop();            ans += x;                 if( a.empty() ) break;            a.push(x);        }        printf("%lld\n", a.empty() ? ans : a.top() );    }    return 0;}