Fence Repair
🖋️ 문제
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.
입력
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
출력
Line 1: One integer: the minimum amount of money he must spend to make N-1 cuts
🤔 접근
처음에 코드부터 치려고 했다가 시간만 날렸다. 결국 종이에 여러 case들을 직접 해보고 어떻게 풀어야 할지 감이 잡혔다.
처음에는 어떻게 나눠가야 최소의 비용이 들지 고민해보았지만 규칙성을 도저히 찾을 수 없었다. 결국은 어떻게 나누는지가 중요한 것이 아니라 어떤 것을 나누는지가 중요하기 때문이었다. 따라서 거꾸로 판자를 붙여가며 만들어가는 식으로 생각해야 하는 문제였다.
여러 개의 판자 중 가장 작은 두 개를 붙이고 다시 전체 중에서 가장 작은 두 개를 붙여나가는 식으로 해결하였다.
💻 내 코드
import sys
import heapq
input = sys.stdin.readline
n = int(input())
data = [int(input()) for _ in range(n)]
rst = 0
heapq.heapify(data)
while len(data) > 1:
a = heapq.heappop(data)
b = heapq.heappop(data)
heapq.heappush(data, a+b)
rst += (a+b)
print(rst)👍 제출
