Algorithm89 [백준] 11725번 트리의 부모찾기 from collections import deque import sys N = int(input()) ans = [0] * (N + 1) mapData = [[] for _ in range(N + 1)] for i in range(N - 1): V1, V2 = map(int, sys.stdin.readline().split()) mapData[V1].append(V2) mapData[V2].append(V1) que = deque([1]) visited = [0] * (N + 1) while que: now = que.popleft() for i in mapData[now]: if not visited[i]: ans[i] = now que.append(i) visited[i] = 1 for i in a.. 2021. 2. 8. [백준] 11725번 트리의 부모찾기 from collections import deque import sys N = int(input()) ans = [0] * (N + 1) mapData = [[] for _ in range(N + 1)] for i in range(N - 1): V1, V2 = map(int, sys.stdin.readline().split()) mapData[V1].append(V2) mapData[V2].append(V1) que = deque([1]) visited = [0] * (N + 1) while que: now = que.popleft() for i in mapData[now]: if not visited[i]: ans[i] = now que.append(i) visited[i] = 1 for i in a.. 2021. 2. 8. [백준] 18352번 특정 거리의 도시 찾기 from collections import deque import sys N, M, K, X = map(int, sys.stdin.readline().split()) visited = [-1] * (N + 1) visited[X] = 0 mapData = [[] for _ in range(N + 1)] for i in range(M): V1, V2 = map(int, sys.stdin.readline().split()) mapData[V1].append(V2) que = deque([X]) while que: now = que.popleft() for nxt in mapData[now]: if visited[nxt] == -1: visited[nxt] = visited[now] + 1 que.append.. 2021. 2. 8. [백준] 10819번 차이를 최대로 from itertools import permutations를 활용하면 성능은 좀 그렇지만 쉽게 구현이 가능하다.. from itertools import permutations lengthOfData = int(input()) dataList = list(map(int, input().split())) ans = 0 for i in permutations(dataList): tempVal = 0 for j in range(lengthOfData - 1): tempVal += abs(i[j] - i[j + 1]) ans = max(ans, tempVal) print(ans) 2021. 2. 5. 이전 1 2 3 4 5 ··· 23 다음