POJ 3253 Fence Repair

Fence Repair

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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
    
    
    
    
    
    
    
    

     
     
     
     8
    
    
    
    
    
    
    
    

     
     
     
     5
    
    
    
    

Sample Output

34

大致题意：农夫约翰要将一块很长的木板切割成n块，准备切成的木板的长度为L1,L2.....LN,未切割木板的长度恰好为切割后木板长度的总和。每次切断木板时，需要的开销为这块木板的长度。

分析：此题可以转化为哈夫曼树的构造问题，用固定长度的Li（1<= i <= N）：（1）从N个木块中每次找出最短的两块，合并，放入队列中（2）重复1过程，知道队列中剩余一个元素。如此，便可保证总消耗最小。

<pre name="code" class="cpp">#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<queue>

using namespace std;
//重载操作，将队列中的元素从小到大排列
struct cmp
{
    bool operator()(int &a,int &b)
    {
        return a > b;
    }
};

int main()
{
    int t;
    while(~scanf("%d",&t))
    {
        priority_queue<int,vector<int>,cmp>que;//定义优先队列
        int x;
        for(int i = 0;i < t;i++)
        {
            scanf("%d",&x);
            que.push(x);
        }
        long long sum = 0;
        while(que.size() > 1)
        {
            int a = que.top();
            que.pop();
            int b = que.top();
            que.pop();
            sum += a + b;
            que.push(a + b);
        }
        printf("%lld\n",sum);
    }
    return 0;
}