Fence Repair-贪心优先队列

题目：

    

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.

输入：

    Line 1: One integer <i>N</i>, the number of planks<br>Lines 2..<i>N</i>+1: Each line contains a single integer describing the length of a needed plank

输出：

    Line 1: One integer: the minimum amount of money he must spend to make <i>N</i>-1 cuts

样例：

输入：

3
8
5
8输出：34题目大意：一个木板，截成n块，截多少花多少钱，求花钱最少。思路：用优先队列，每次得出并删除木板要截出的最小值和次最小值，再将这两个的和加入队列，截n-1次得出结果。代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<queue>
using namespace std;
int n,a;
long long ans=0;

int main()
{
    cin>>n;
    priority_queue <int,vector<int>,greater<int> > q;
    for(int i=1;i<=n;i++)
    {
        cin>>a;
        q.push(a);
    }
    while(n>1)
    {
        int t1=q.top();
        q.pop();
        int t2=q.top();
        q.pop();
        ans+=t1+t2;
        q.push(t1+t2);
        n--;
    }
    cout<<ans<<endl;
    return 0;
}