[프로그래머스 LV 2] 배달 파이썬/python

https://programmers.co.kr/learn/courses/30/lessons/12978

코딩테스트 연습 - 배달

 

플로이드 워셜과 다익스트라 두가지로 풀어봄

그래프 최단경로 문제, 특히 다익스트라랑 플로이드워셜을 처음 써봐서 조금 헤맴

근데 한 세시간정도 코드랑 씨름하다보니까 알고리즘 기전은 확실히 알게된듯

 

 

플로이드워셜

def solution(N, road, K):
    INF = 1000000000
    answer = 0
    graph = [[INF] * (N + 1) for _ in range(N + 1)]
 
    for i in range(1, N + 1):
        for j in range(1, N + 1):
            if i == j:
                graph[i][j] = 0
 
    for r in road:
        a, b, c = r[0], r[1], r[2]
        if graph[a][b] > c:
            graph[a][b] = c
            graph[b][a] = c
 
    for i in range(1, N + 1):
        for j in range(1, N + 1):
            for k in range(1, N + 1):
                graph[j][k] = min(graph[j][k], graph[j][i] + graph[i][k])
 
    for j in range(1, N + 1):
        if graph[1][j] <= K:
            answer += 1
 
    return answer
 
 
 

 

 

다익스트라

from queue import PriorityQueue
 
 
def solution(N, road, K):
    INF = 1000000000
    answer = 0
    distance = [INF] * (N + 1)
 
    q = PriorityQueue()
    q.put([1, 0])
    distance[1] = 0
    while not q.empty():
        now, now_cost = q.get()
 
        if now_cost > distance[now]: continue
        for start, dest, cost in road:
            next_cost = now_cost + cost
            if start == now and next_cost < distance[dest]:
                distance[dest] = next_cost
                q.put([dest, next_cost])
            elif dest == now and next_cost < distance[start]:
                distance[start] = next_cost
                q.put([start, next_cost])
 
    for i in range(1, N + 1):
        if distance[i] <= K:
            answer += 1
 
    return answer