POJ 3253 Fence Repair

最新推荐文章于 2023-03-04 18:57:46 发布

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本文介绍了一种解决农民约翰修理栅栏所需最小化锯木成本的算法。通过从大到小排序木板长度并进行最优切割，农民约翰能够以最低成本获得所需的木板数量。

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                                              Fence Repair
                               Time Limit: 2000MS       Memory Limit: 65536K

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).

题目描述：得到n个长度分别为ai的木棍。例如样例中的5 8 8.开始我们有一根长为21的木棍，第一次截断，得到长为13和8的木棍，代价为21。第二次截断得到长为8 和5 的木棍，代价为13，然后总共的最小代价为34，输出答案。
算法设计：对n个木棍的长度进行从大到小排序。然后开始合并第n个木棍和第n-1个木棍（最小的，和次小的）。记录合并的代价，然后将新得到的木棍（长度为两木棍之和）“插排”进数组，使数组中的数依然有序。然后继续执行上述操作，直至只剩下一个木棍。
算法分析：题目利用贪心的思想，递归的对数据进行处理。

#include <cstdio>
#include <algorithm>
#include <iostream>
using namespace std;
long long a[20010];
long long b[20010];
int main()
{
    long long n,sum,ans;
    long long s,p;
    while (scanf("%I64d",&n)==1)
    {
        ans=0;
        for (int i=0;i<n;i++)
        {
            scanf("%I64d",&a[i]);
        }
        sort(a,a+n);
        for (int i=0;i<n;i++)
            b[i]=a[n-i-1];
        /*
        for (int i=0;i<n;i++)
            printf("%d\n",b[i]);
            */
        while (n>1)
        {
            s=b[n-1]+b[n-2];
            ans+=s;
            int i;
            p=0;
            for(i=n-3;i>=0;i--)
            {
                if (s<b[i])
                {
                    p=i;
                    break;
                }
            }
            for (i=n-2;i>p;i--)
                b[i]=b[i-1];
            if (p==0)
              b[0]=s;
            else
              b[p+1]=s;
            n--;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}