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.

#include<stdio.h>

#include<iostream>

#include<queue>

using namespace std;

int main()

{

    priority_queue<int,vector<int>,greater<int> >wood;

    int n;

    cin>>n;

    for(int i=;i<n;i++)

    {

        int tmp;

        cin>>tmp;

        wood.push(tmp);

    }

    long long ans=;

    while(!wood.empty())

    {

        int l1=wood.top();

        wood.pop();

        if(wood.empty()) break;

        ans+=l1;

        int l2=wood.top();

        ans+=l2;

        wood.pop();

        int newone=l1+l2;

        //cout<<l1<<"  "<<l2<<endl;

        wood.push(newone);

    }

    printf("%I64d\n",ans);

    return ;

}

这里注意优先队列对已定义的数据类型的排列，默认是大顶的，变成小顶用到：

priority_queue<int,vector<int>,greater<int> >wood;

自定义优先级：

struct node

{

    friend bool operator< (node n1, node n2)

    {

        return n1.priority < n2.priority;

    }

    int priority;

    int value;

};