网站建设科技风/广东vs北京首钢

网站建设科技风,广东vs北京首钢,免费二级域名注册平台,广州建筑装饰集团有限公司Bellman_ford 队列优化算法（又名SPFA） 代码随想录 import collectionsdef main():n, m map(int, input().strip().split())edges [[] for _ in range(n 1)]for _ in range(m):src, dest, weight map(int, input().strip().split())edges[src].append…

Bellman_ford 队列优化算法（又名SPFA）

代码随想录

import collectionsdef main():n, m = map(int, input().strip().split())edges = [[] for _ in range(n + 1)]for _ in range(m):src, dest, weight = map(int, input().strip().split())edges[src].append([dest, weight])minDist = [float("inf")] * (n + 1)minDist[1] = 0que = collections.deque([1])visited = [False] * (n + 1)visited[1] = Truewhile que:cur = que.popleft()visited[cur] = Falsefor dest, weight in edges[cur]:if minDist[cur] != float("inf") and minDist[cur] + weight < minDist[dest]:minDist[dest] = minDist[cur] + weightif visited[dest] == False:que.append(dest)visited[dest] = Trueif minDist[-1] == float("inf"):return "unconnected"return minDist[-1]if __name__ == "__main__":print(main())

bellman_ford之判断负权回路

代码随想录

import sysdef main():input = sys.stdin.readdata = input().split()index = 0n = int(data[index])index += 1m = int(data[index])index += 1grid = []for i in range(m):p1 = int(data[index])index += 1p2 = int(data[index])index += 1val = int(data[index])index += 1# p1 指向 p2，权值为 valgrid.append([p1, p2, val])start = 1  # 起点end = n    # 终点minDist = [float('inf')] * (n + 1)minDist[start] = 0flag = Falsefor i in range(1, n + 1):  # 这里我们松弛n次，最后一次判断负权回路for side in grid:from_node = side[0]to = side[1]price = side[2]if i < n:if minDist[from_node] != float('inf') and minDist[to] > minDist[from_node] + price:minDist[to] = minDist[from_node] + priceelse:  # 多加一次松弛判断负权回路if minDist[from_node] != float('inf') and minDist[to] > minDist[from_node] + price:flag = Trueif flag:print("circle")elif minDist[end] == float('inf'):print("unconnected")else:print(minDist[end])if __name__ == "__main__":main()

bellman_ford之单源有限最短路

代码随想录

def main():# 輸入n, m = map(int, input().split())edges = list()for _ in range(m):edges.append(list(map(int, input().split() )))start, end, k = map(int, input().split())min_dist = [float('inf') for _ in range(n + 1)]min_dist[start] = 0# 只能經過k個城市，所以從起始點到中間有(k + 1)個邊連接# 需要鬆弛(k + 1)次for _ in range(k + 1):update = Falsemin_dist_copy = min_dist.copy()for src, desc, w in edges:if (min_dist_copy[src] != float('inf') and min_dist_copy[src] + w < min_dist[desc]):min_dist[desc] = min_dist_copy[src] + wupdate = Trueif not update:break# 輸出if min_dist[end] == float('inf'):print('unreachable')else:print(min_dist[end])if __name__ == "__main__":main()

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