POJ 3253 Fence Repair

本文介绍了一种通过优先队列实现的最优木板切割算法,旨在帮助农夫以最低成本将长木板切割成所需长度的小木板。该算法利用了贪心策略,确保每次切割都是成本最低的选择。

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Fence Repair
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 51663 Accepted: 16974

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

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).

题目大意: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

优先队列水题,每次算最小的即可,要用long long
#include<iostream>
#include<queue>
using namespace std;
typedef long long ll;
int main(){
	ios::sync_with_stdio(0);
	int n;
	cin>>n;
	priority_queue<int,vector<int>,greater<int> >que;
	while(n--){
		int x;
		cin>>x;
		que.push(x);
	}
	ll ans=0;
	while(!que.empty()){
		int a=que.top();
		ans+=a;
		que.pop();
		if(que.empty()) break;
		int b=que.top();
		ans+=b;
		que.pop();
		if(que.empty()) break;
		que.push(a+b);
	}
	cout<<ans<<endl;
	return 0;
} 




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