http://poj.org/problem?id=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
Lines 2..N+1: Each line contains a single integer describing the length of a needed plank
Output
Sample Input
3
8
5
8
Sample Output
34
Hint
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).
这个题说的是一个人要围篱笆，有一块木板，把它切成n份，每份i米长，而没切割一次就收切割成两份的总长的价格，比如8米的木板切成3，5，就要付8美元，然后问你最少付多少钱
然后我想的就是把序列从小到大排列，因为要求最小的钱，所以最大的木板切一次就够了，小的切几次也没关系，所以先把所有加一遍，然后n-1个加起来，再把n-2ge加起来，……最后把所有的都加起来，
但是书上说把它看成树，有多少层就切割多少次，所以比较小的要在最小层，先排序，之后把最小的两个相加，把它当作一个板子，这样就相当于有n-1个板子，依次……
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
int main()
{
int n,a[50000];
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
ll sum=0;
while(n>1)//直到计算出短板为1
{
int l1=0,l2=1;
if(a[l1]>a[l2])
{
swap(l1,l2);
}
for(int i=2;i<n;i++)
{
if(a[i]<a[l1])
{
l2=l1;
l1=i;
}
else if(a[i]<a[l2])
{
l2=i;
}
}
int s=a[l1]+a[l2];
sum+=s;
if(l1==n-1)
swap(l1,l2);
a[l1]=s;
a[l2]=a[n-1];
n--;
}
printf("%lld\n",sum);
return 0;
}