sdutacm-Fence Repair

最新推荐文章于 2018-12-11 18:08:00 发布

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最新推荐文章于 2018-12-11 18:08:00 发布

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分类专栏： ACM《数据结构》树 - 堆、哈夫曼树文章标签： acm sdut c语言算法

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FenceRepair

TimeLimit: 2000MS Memory Limit: 65536KB

SubmitStatistic

Problem Description

Farmer John wants to repair a small length of the fencearound the pasture. He measures the fence and finds that he needs N (1≤ N ≤ 20,000) planks of wood, each having some integerlength L_i (1 ≤ L_i ≤ 50,000)units. He then purchases a single long board just long enough to saw intothe N planks (i.e., whose length is the sum of thelengths L_i). FJ is ignoring the "kerf", theextra 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 whichto cut the wood, so he mosies over to Farmer Don\'s Farm with this long boardand politely 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 the N-1 cutsin 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 andlocations to cut the plank. Help Farmer John determine the minimum amount ofmoney he can spend to create the N planks. FJ knows that hecan cut the board in various different orders which will result in differentcharges 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-1cuts

Example Input

Example Output

Hint

Author

#include <iostream>
#include<string.h>
#include<stdio.h>
#include<queue>
#include<stdlib.h>
#include<algorithm>

using namespace std;
long long sum;
int h[10002];

int main()
{
    priority_queue<int,vector<int>,greater<int> >q;
    int n,i,a,b,c;
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        scanf("%d",&h[i]);
        q.push(h[i]);
    }
    sum = 0;
    while(!q.empty())
    {
        a = q.top();
        q.pop();
        if(!q.empty())
        {
           b = q.top();
           q.pop();
           c = a+b;
           q.push(c);
           sum += c;
        }
    }
    printf("%lld\n",sum);
    return 0;
}
/*题目大意：FJ需要修补牧场的围栏，他需要 N 块长度为 Li 的木头（N planks of woods）。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，Farmer Don提供FJ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，FJ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


题目可以转化为Huffman树构造问题 ：

给定 N planks of woods，

1. 　　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，

2.　　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
*/

/***************************************************
User name: jk160505徐红博
Result: Accepted
Take time: 12ms
Take Memory: 428KB
Submit time: 2017-02-09 09:41:23
****************************************************/