代码测试

import heapq
import sys
import math

sys.setrecursionlimit(10000)
INF = 0x3f3f3f3f

## 定义类
class Point:
    def __init__(self, x, y):
        self.x = float(x)
        self.y = float(y)

# 计算欧式距离
def get_distance(p1, p2):
    return math.sqrt(abs(p1.x - p2.x) ** 2 + abs(p1.y - p2.y) ** 2)


def dijkstra(G, start, endp):
    dis = dict((point, INF) for point in G)
    dis[start] = 0.00
    vis = dict((point, False) for point in G)
    # heap
    pq = []
    heapq.heappush(pq, [dis[start], start])
    path = dict((point, [start]) for point in G)
    while len(pq) > 0:
        v_dis, v = heapq.heappop(pq)
        if vis[v] == True:
            continue
        vis[v] = True
        p = path[v].copy()
        for point in G:
            distance = get_distance(v, point)
            if distance > 100.00:
                continue
            new_dis = dis[v] + distance

            if new_dis < dis[point] and (not vis[point]):
                dis[point] = new_dis

                heapq.heappush(pq, [dis[point], point])
                temp = p.copy()
                temp.append(point)
                path[point] = temp
    distance = dis[endp]
    if distance == INF:
        print("-1")
    else:
        print("{:.2f}".format(distance))


while True:
    try:
        numbers = input().strip().split(",") # 根据','进行分割
        limit = int(numbers[0])
        point_list = []
        numbers = list(map(float, numbers))
        stap = Point(numbers[1], numbers[2])
        endp = Point(numbers[-2], numbers[-1])
        if stap.x > limit or stap.y > limit or stap.x < 0 or stap.y < 0:
            print("-1")
        elif endp.x > limit or endp.y > limit or endp.x < 0 or endp.y < 0:
            print("-1")
        else:
            for i in range(1, len(numbers), 2):
                if numbers[i] <= limit and numbers[i + 1] <= limit:
                    point_list.append(Point(numbers[i], numbers[i + 1]))
            dijkstra(point_list, point_list[0], point_list[-1])
    except EOFError:
        break

标签：distance,pq,point,代码,numbers,测试,endp,dis
来源： https://www.cnblogs.com/CharlesLC/p/15431685.html