Fence Repair POJ - 3253
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).
Code
/*
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*/
/* Procedural objectives：
Variables required by the program:
Procedural thinking：
如果用二叉树对应切割方法，每一个叶子节点就对应了切割出的一块块木板。
叶子节点的深度就对应了为了得到对应木板所需的切割次数，开销的合计就是各叶子节点的 木板的长度*节点的深度。
于是，此时的最佳切割方法首先应该：最短的板与次短的板的节点应当是兄弟节点。
对于最优解来讲，最短的板应当是深度最大的叶子节点之一。
所以与这个叶子节点同一深度的兄弟节点一定存在，并且由于同样是最深的叶子节点，所以应该对应于次短的板。
由于只需从板的集合里取出最短的两块，并且把长度为两块板长度之和的板加入集合中即可，因此如果使用优先队列就可以高效的实现。
Functions required by the program:
Determination algorithm:
Determining data structure:
*/
/* My dear Max said：
"I like you,
So the first bunch of sunshine I saw in the morning is you,
The first gentle breeze that passed through my ear is you,
The first star I see is also you.
The world I see is all your shadow."
FIGHTING FOR OUR FUTURE！！！
*/
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
int N,L[20007];
void solve(){
long long ans=0;
priority_queue< int,vector<int>,greater<int> > que;
for(int i=0;i<N;i++)
que.push(L[i]);
while(que.size()>1){
int l1,l2;
l1=que.top();
que.pop();
l2=que.top();
que.pop();
ans+=l1+l2;
que.push(l1+l2);
}
cout<<ans<<endl;
}
int main(){
cin>>N;
for(int i=0;i<N;i++)
cin>>L[i];
solve();
return 0;
}
