POJ 3253 Fence Repair (哈夫曼)

Fence Repair

 

Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 45098 Accepted: 14700

 

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 Li (1 ≤ Li ≤ 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 Li). 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

 

题意

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,求最小的花费。

比如说测试数据 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。

 

思路

既然最终要找的是长度为 Li 的 N 块木板,那么我们可以用从下向上组合的方式选择(类似于数塔问题)

首先假设我们有长度为5 8的两根木头,那它是由长度为13的木头得来的,于是,此时便有了长13 8的两根木头,同样,它是由长21的木头得来的,于是最终的最小花费为 13+21=34

哈夫曼问题,总是选取最小的两个,然后组合之后加入队列,当到达树的顶端时,组合完毕,输出路径中各组合节点的和。

 

AC 代码

#include <iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<map>
#include<queue>
#include<algorithm>
using namespace std;
typedef __int64 LL;

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        priority_queue<int,vector<int>,greater<int> >sk;    //优先队列 小顶堆
        int num;
        LL ans=0;
        for(int i=0; i<n; i++)
        {
            scanf("%d",&num);
            sk.push(num);
        }
        if(sk.size()==1)    //如果只有1根木头
            ans=sk.top();
        while(sk.size()>1)
        {
            int a=sk.top();
            sk.pop();
            a+=sk.top();    //选取最小的两个求和
            ans+=a;
            sk.pop();
            sk.push(a); //加入队列
        }
        printf("%I64d\n",ans);
    }
    return 0;
}