POJ 3253 Fence Repair 贪心 优先级队列

Fence Repair

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 77001		Accepted: 25185

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

 

之前我们用数组实现了这个贪心，简单且很容易就AC了

但是，还有更简单，并且更快的写法

那就是优先级队列，直接调用STL实现即可

 

 #include <cstdio>
 #include <iostream>
 #include <queue>

 using namespace std;

 const int max_n=;
 const int max_L=;

 typedef long long LL;

 int n;
 int L[max_n];

 void solve()
 {
     LL ans=;
     // 建立最小优先队列
     priority_queue< int,vector<int>,greater<int> > que;
     for(int i=;i<n;++i)
     {
         que.push(L[i]);
     }

     int tmp;
     while(que.size()>=)
     {
         tmp=que.top();
         que.pop();
         tmp+=que.top();
         que.pop();

         ans+=tmp;
         que.push(tmp);
     }

     printf("%lld",ans);
 }

 int main()
 {
     scanf("%d",&n);
     for(int i=;i<n;++i)
     {
         scanf("%d",&L[i]);
     }
     solve();
     return ;
 }

嗯，代码和行数只有之前的一半，不经思考直接调库还不损耗脑细胞。

关键是，更快了

 

21294701

LIUYUANHAO

3253

Accepted

220K

844MS

C++

1741B

2020-02-02 18:11:44

21310136

LIUYUANHAO

3253

Accepted

344K

0MS

C++

665B

2020-02-05 17:31:23

 

复杂度从o（n^2) 降到了 o（ n*log(n) )