Fence Repair oj

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本文介绍了一种帮助农民John通过最优切割策略最小化成本的方法。面对需要从一块长木板上切割出N块特定长度木板的问题，文章提供了一个算法解决方案，通过构建数据结构来跟踪切割过程并确保每次切割成本最低。

Fence Repair

Time Limit: 2000MS Memory Limit: 65536KB

Problem 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

Example Input

Example Output

Hint

Author

#include <stdio.h>
#include <stdlib.h>
#define Max 30000
int a[Max];
struct node
{
    int weight;
    int parent,l,r;
}L[Max];

int main()
{
    int n,m,i,j;
    scanf("%d",&n);
    m = 2*n-1;
//    printf("%d\n", (1<<31-1));
    for(i = 1;i <= n;i++)
    {
        scanf("%d",&L[i].weight);
    }
    for(i = 1;i <= m;i++)
    {
        L[i].parent =  L[i].l = L[i].r = 0;
    }
    for(i = n+1;i<=m;i++)
    {
        int p1,p2,s1,s2;
        p1 = p2 = 0,s1 = s2 =(1<<31-1);
        for(j = 1;j <= i-1;j++)
        {
            if(L[j].parent == 0)
            {
                if(L[j].weight < s1)
                {
                    p2 = p1;
                    p1 = j;
                    s2 = s1;
                    s1 = L[j].weight;
                }
                else if(L[j].weight < s2)
                {
                    s2 = L[j].weight;
                    p2 = j;
                }
            }
        }
        L[p1].parent = i;
        L[p2].parent = i;
        L[i].l = p1;
        L[i].r = p2;
        L[i].weight = L[p1].weight + L[p2].weight;
    }
    long long int num,sum;
    num = sum = 0;
    for(i = 1;i <= n;i++)
    {
        num = 0;
        j = i;
        while(L[j].parent != 0)
        {
            num++;
            j = L[j].parent;
        }
        sum += L[i].weight * num;
    }
    printf("%lld\n",sum);
    return 0;
}