Fence Repair POJ

(英文不好的同学可以参考白书)

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).

 

 

题解：

　　这个题目来自于白书，而且题解也很详细，但为了方便没有白书的同学，这里还是讲一下。

　　首先，我们可以思考将一个大区间不断地切成两瓣，直到切到我们想要的节点为止，可以得到一棵类似与线段树的二差树，然后每个叶子节点就是题目所给的区间，这个可以自己在脑中想一下。

　　然后，我们考虑，对于每个叶子节点，他的花费就是　长度＊深度。因为深度就是切割的次数，然后可以用叶子节点拼凑出大的区间，乘法结合率一下就证明了，那么肯定长度最小的的节点放在深度最大的地方，因为他是一棵二差树，所以次小的区间也放在深度最大的地方，这样一定是最优的，然后考虑将他们和起来，得到划分前的区间形态，然后又是一样的方法找最小次小，知道区间个数为１。

　　用堆维护就一下就好了。

 

代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <iostream>
#include <queue>
#define MAXN 200000
#define ll long long
using namespace std;
int n,tot=0;
ll ans=0;
priority_queue<int> q;

bool cmp(int x,int y){
    return x>y;
}

int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        int x;scanf("%d",&x);
        q.push(-x);
    }
    while(q.size()!=1){
        int f=q.top();q.pop();
        int s=q.top();q.pop();
        q.push(s+f);
        ans+=-s-f;
    }
    printf("%lld
",ans);
    return 0;
}