POJ 3253 优先队列

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

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·：从一块无限长的木板上截取n块给定长度的小木板 每次的费用为所截取的木板长度 求最小的费用

这个题逆向思维  用贪心 给你了n块木板的长度 可以把这n块木板拼接成一块木板 每次拼接两块 只要每次拼接时花费的费用是最小的 那么最终截取的时候总费用也是最少的

虽然思路是这样的 但是我直接暴力超时了 于是去网上搜了一下别人的题解 发现可以用哈曼夫树 和 优先队列

哈曼夫树用的不是很熟 就去看了一下优先队列 发现优先队列很简单

第一次数据开小了 wa了一次  改成long long 就过了.....

ac代码：

#include
#include
#include
#include
#include

using namespace std;

struct cmp
{
    bool operator()(long long x,long long y)
    {
        return x > y;
    }
};




int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        priority_queue,cmp>q;

        for(int i = 0; i < n; i++)
        {
            long long temp;
            scanf("%lld",&temp);
            q.push(temp);
        }

        long long sum = 0;
        while(q.size() > 1)
        {
            long long a = q.top();
            q.pop();
            long long b = q.top();
            q.pop();
            long long c = a+b;
            q.push(c);
            sum += c;
        }

        printf("%lld\n",sum);
    }
    return 0;
}