POJ 3253 贪心法，优先队列

Fence Repair

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 35249 Accepted: 11392

Description

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

3858

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).

遇到的问题和思路：

       首先吧，看到这道题真心不会。因为优先队列的知识没有学过，所以做起来有点吃力。起初看了题解以为只是单纯的排序好放进去就好了，根本就没有想到有优先队列，后来才知道，原来优先队列还是有自动排序的，长知识了，有收获！

给出AC代码：

#include<cstdio>#include<algorithm>#include<cstring>#include<cmath>#include<queue>using namespace std;int a[20005];int t;long long ans;//首先注意queue的（）不能忘了，然后通过ans来计数不能忘了。还有一定要先对a数组进行排序 void solve(){priority_queue <int, vector<int>, greater<int> > que;ans = 0; for (int i = 0; i < t; i++)    que.push(a[i]);while (que.size() > 1){int a1 = que.top();que.pop();int a2 = que.top();que.pop();que.push(a1 + a2);ans = ans + a1 + a2;}printf("%I64d\n",ans);}int main(){while(scanf("%d",&t)!=EOF){memset(a, 0, sizeof(a));for(int i = 0; i < t; i++){scanf("%d",&a[i]);}sort(a,a+t);solve();}return 0;}