POJ 3253

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本文介绍了一种通过算法确定最优切割顺序的方法，旨在帮助农民John以最低成本将长木板切割成所需尺寸的小木板。文章详细解释了如何通过寻找最短及次短木板并进行合并的方式，来最小化总切割费用。

Fence Repair

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 40620		Accepted: 13291

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

<span style="font-family:Courier New;font-size:18px;">#include <iostream>
#include <cstdio>
#include <math.h>
using namespace std;

#define MAX_N 50010
typedef long long ll;
int N, L[MAX_N];

void Input()
{
    scanf("%d", &N);
    for(int i = 0; i < N; i++)
        scanf("%d", &L[i]);
}

void Solve()
{
    ll ans = 0;

    while(N > 1) {
        ///  求出最短的板mii1和次短的板mii2
        int mii1 = 0, mii2 = 1;
        if (L[mii1] > L[mii2]) swap(L[mii1], L[mii2]);

        for(int i = 2; i < N; i++) {
            if (L[i] < L[mii1]) {
                mii2 = mii1;
                mii1 = i;
            }
            else if (L[i] < L[mii2]) {
                mii2 = i;
            }
        }

        int t = L[mii1] + L[mii2];
        ans += t;

        if(mii1 == N - 1) swap(mii1, mii2);
        L[mii1] = t;
        L[mii2] = L[N - 1];
        N--;
    }

    printf("%lld\n", ans);
}

int main()
{
    Input();
    Solve();
    return 0;
}
/*
3
8
5
8
8
*/</span>