poj 3253 Fence Repair

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本文介绍了一种基于赫夫曼思想的最优切割策略，通过使用优先队列来解决农夫约翰需要将一块长木板切割成多段的问题，旨在找到成本最低的切割方案。

Fence Repair

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 40272		Accepted: 13156

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

Sample Output

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).

Source

USACO 2006 November Gold

5 8 5为例：

先从无限长的木板上锯下长度为 21 的木板，花费 21

再从长度为21的木板上锯下长度为5的木板，花费5

再从长度为16的木板上锯下长度为8的木板，花费8

总花费 = 21+5+8 =34

解题思路：

利用Huffman思想，要使总费用最小，那么每次只选取最小长度的两块木板相加，再把这些“和”累加到总费用中即可

本题虽然利用了Huffman思想，但是直接用HuffmanTree做会超时，可以用优先队列做

用优先队列做就可以了

记住一定要用long long 啊！

#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <queue>
#include <iostream>
using namespace std;

int n, m;
int a[20005];
struct cmp1
{
    bool operator ()(int &a,int &b)
    {
        return a>b;                                          //最小值优先
    }
};
priority_queue<int,vector<int>,cmp1> q;

void init ()
{
    while ( !q.empty() )
    {
        q.pop();
    }
}

long long Pri ()
{
    int x, y;
    long long sum = 0;
    
    while ( q.size() != 1 )
    {
        x = q.top();
        q.pop();
        y = q.top();
        q.pop();
        sum = (long long)(x+y)+sum;
        x = x+y;
        q.push(x);
    }
    
    return sum;
}

int main()
{
    int i;
    scanf ( "%d", &n ) ;
    
    /*Intialize the data  初始化数据*/
    init();

    /*Storage of the data  数据储存*/
    for ( i = 0; i < n; i++ )
    {
        scanf ( "%d", &a[i] );
        q.push ( a[i] );
    }

    /*Priority queue output  优先队列的输出*/
    long long sum = Pri ();
    printf ( "%lld\n", sum );
}

代码菜鸟，如有错误，请多包涵！！！