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 Nplanks. 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.
Lines 2.. N+1: Each line contains a single integer describing the length of a needed plank
3 8 5 8
34
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】
其实刚开始是想用区间dp做的，后来推公式的时候发现有规律，就直接用优先队列了，题意就是，切一个无限长的篱笆，告诉你要切的长度，比如样例，先切个21，花费21，21被切成13和8 ，再把13切成8和5，花费13，所以总共花费34，要你求最小花费。
其实规律就是每次在一堆长度里找到最短的和次短的，相加后放入队列中，再次找到最短和次短的，一直重复，直到只剩下一个长度为止。
优先队列裸题。
要注意用long long ;
【AC代码】
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f;
int m;
ll sum;int main()
{ll a;while(~scanf("%d",&m)){priority_queue<ll, vector<ll>, greater<ll> >q;//大顶堆while(!q.empty()) q.pop();for(int i=1;i<=m;++i){scanf("%lld",&a);q.push(a);}sum=0;if(q.size()==1){a+=q.top();sum+=a;q.pop();}else{ll st,ed;while(q.size()>1){st=q.top();q.pop();ed=q.top();q.pop();ll s=ed+st;sum+=s;q.push(s);}printf("%lld\n",sum);}}return 0;
}