POJ 3253 优先队列实现哈弗曼树

Fence Repair

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 40009		Accepted: 13044

Description

Farmer John wants to repair a small length of the fencearound the pasture. He measures the fence and finds that he needsN (1 ≤N ≤ 20,000) planks of wood, each having some integer lengthL_i(1 ≤ L_i ≤ 50,000) units. He then purchases a single longboard just long enough to saw into theN planks (i.e., whose length isthe sum of the lengths L_i). FJ is ignoring the"kerf", the extra length lost to sawdust when a sawcut is made; youshould ignore it, too.

FJ sadly realizes that he doesn't own a saw with which tocut the wood, so he mosies over to Farmer Don's Farm with this long board andpolitely asks if he may borrow a saw.

Farmer Don, a closet capitalist, doesn't lend FJ a sawbut instead offers to charge Farmer John for each of theN-1 cuts in theplank. 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 andlocations to cut the plank. Help Farmer John determine the minimum amount ofmoney he can spend to create theN planks. FJ knows that he can cut theboard in various different orders which will result in different charges sincethe 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 lengthof a needed plank

Output

Line 1: One integer: the minimum amount of money hemust spend to makeN-1 cuts

Sample Input

3

8

5

8

Sample Output

34

Hint

He wants to cut a board of length 21 into pieces oflengths 8, 5, and 8.
The original board measures 8+5+8=21. The first cut will cost 21, and should beused to cut the board into pieces measuring 13 and 8. The second cut will cost13, and should be used to cut the 13 into 8 and 5. This would cost 21+13=34. Ifthe 21 was cut into 16 and 5 instead, the second cut would cost 16 for a totalof 37 (which is more than 34).

题目大意：农民需要修补牧场的围栏，他需要 N 块长度为 Li 的木头。开始时，FJ只有一块无限长的木板，因此他需要把无限长的木板锯成 N 块长度为 Li 的木板，另一个农民提供锯子，但必须要收费的，收费的标准是对应每次据出木块的长度，比如说测试数据中 5 8 8，一开始，需要在无限长的木板上锯下长度 21 的木板（5+8+8=21），第二次锯下长度为 5 的木板，第三次锯下长度为 8 的木板，至此就可以将长度分别为 5 8 8 的木板找出，费用为21+5+8=34.

 

思路：用哈弗曼树（最优二叉树）的思想的话就可以看成是把n段木板拼成一块，每次拼的代价为所拼木板的长度和。每次选取两个最小的来拼。在 N planks 中每次找出两块长度最短的木板，然后把它们合并，加入到集合A中，在集合中找出两块长度最短的木板，合并加入到集合A中，重复过程，直到集合A中只剩下一个元素

显然，通过每次选取两块长度最短的木板，合并，最终必定可以合并出长度为 Sum（Li）的木板，并且可以保证总的耗费最少。直接用STL的优先队列做非常方便，声明一个最大优先队列，每次取出最队列最小的值。看了discuss提示到测试数据是存在大数的情况的，要使用long long长整形。

以前一直没有在输入处加不等于EOF的习惯，这题偏偏不加就会超时。。。。。！以前就没碰到过，习惯还是要养好。。。。。。。

#include<cstdio>
#include<queue>
using namespace std;

priority_queue<int,vector<int>,greater<int> >q; //声明最小优先队列

int main()
{
    long long int sum;
    int i,n,s,a,b,tmp;
    while(scanf("%d",&n)!=EOF)
    {
      //把节点都入队列
        for(i=0; i<n; i++)
        {
            scanf("%d",&s);
            q.push(s);
        }
        sum=0;
  
        if(q.size()==1) //如果队列只剩一个点，那就是根，已遍历完
        {
            a=q.top();
            sum+=a;
            q.pop();
        }
        while(q.size()>1)
        {
            a=q.top();//每次取的为最小值（优先级更高）
            q.pop();
            b=q.top();
            q.pop();
            tmp=a+b;
            sum+=tmp;//价格为生成的节点的权值
            q.push(tmp); //把新生成点入队列
        }
        printf("%lld\n",sum);
    }
    return 0;
}