Fence Repair (优先队列)

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).

题意：需要n块木板，每块木板长度为Li，如果切割长度为L的木板需要花费L，求最小花费

开始时一直以为每次切掉最长的木板就可以保证最后花费最小，直到大佬给我举了一个Li依次为1，2，3，4，5的反例，如果先切成6和9，再切成4,5,3,3最后切成4,5,3,2,1，这样总花费为15+9+6+3明显小于我开始以为的那中策略。尝试倒过来思考，既然切成长度为a和b的木板需要(a+b)的花费，那么把a和b拼在一起也需要(a+b)的花费，所以每次都选取最小的两个板，保证每一步花费最小，最后拼成一整块木板也是最小花费，用一个优先队列很容易实现。

#include<cstdio>#include<queue>using namespace std;priority_queue<int, vector<int>, greater<int> > q;int main(){    int n, t, t1, t2;    long long ans = 0;    scanf("%d", &n);    for(int i=0;i<n;i++)    {        scanf("%d", &t);        q.push(t);    }    while(q.size()>=2)    {        t1 = q.top();        q.pop();        t2 = q.top();        q.pop();        ans += t1 + t2;        q.push(t1+t2); //       printf("t=%d", t1+t2);    }    printf("%lld", ans);    return 0;}