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POJ 3253 Fence Repair 类似哈夫曼树的贪心思想

2024-07-09 11:33:31   222人阅读

Fence Repair
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 24550 Accepted: 7878

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

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

题解

用类似哈夫曼树的思想去考虑，每次选取两个较小的点合并，然后把合并后的放回原集合。用一个小根堆去维护即可。

注意首先，C++的优先队列默认是大顶堆，所以需要改写；其次，改写之后的堆是没有clear()方法的，所以务必记住队列的清空！

代码示例

/**============================================================================
#       COPYRIGHT NOTICE
#       Copyright (c) 2014 All rights reserved
#       ----Stay Hungry Stay Foolish----
#
#       @author       :Shen
#       @name         :POJ 3253
#       @file         :G:\My Source Code\【ACM】训练\0624 - 基础\poj3253.cpp
#       @date         :2014-06-24 14:45
#       @algorithm    :Greedy
============================================================================**/

//#pragma GCC optimize ("O2")
//#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <queue>
#include <cstdio>
#include <string>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
using namespace std;
template<class T>inline bool updateMin(T& a, T b){ return a > b ? a = b, 1: 0; }
template<class T>inline bool updateMax(T& a, T b){ return a < b ? a = b, 1: 0; }

typedef long long int64;

int n, l;

void solve()
{
    priority_queue<int, vector<int>, greater<int> > pq;
    for (int i = 0; i < n; i++)
    {
        scanf("%d", &l);
        pq.push(l);
    }
    int64 ans = 0;
    if (n == 1) ans = l;
    while (pq.size() > 1)
    {
        int a = pq.top(); pq.pop();
        int b = pq.top(); pq.pop();
        int c = a + b;
        ans += c; pq.push(c);
    }
    printf("%d\n", ans);
}

int main()
{
    while (~scanf("%d", &n)) solve();
    return 0;
}


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