数据结构与算法 Farmer John 问题 农夫锯木板问题

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.

For example, 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).

Now, please design an algorithm to help FJ with minimum cost.

题目大概是讲有一个农夫要把一个木板锯成几块给定长度的小木板，每次锯都要收取一定费用，这个费用就是当前锯的这个木板的长度，给定小木板的个数n，及各个要求的小木板的长度，求最小费用。

例如长度为21 的木板要切成长度为5、8、8的三块木板。长21的木板切成长为13和8的板时，开销为21，再将长度为13的板切成长度为5和8的板时，开销是13，于是合计开销为34。

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。

#include
#include
#include
using namespace std;

int main()
{
    int *a;
    int i,j,n,sum;
    __int64 ans;   
    while(1)
    {
	    printf("请输入所需木板的数量：");
	    scanf("%d",&n);
        a=new int[n];   //每块木板的长度
        for(i=0;ia[j])
                    a[j-1]=a[j];
                else
                {
                    a[j-1]=sum;
                    break;
                }
            }
            if(j==n)   //说明sum比所有元素都大,把sum直接放在最后
                a[j-1]=sum;
        }
        printf("\n总费用为：%I64d\n\n",ans);	
    }
    return 0;
}