leetcode1334

from heapq import heappush, heappop
class Solution:
    def findTheCity(self, n: int, edges: 'List[List[int]]', distanceThreshold: int) -> int:
        dis = [[float('inf')]*n for _ in range(n)]
        nei = collections.defaultdict(list)
        for i,j,x in edges:
            dis[i][j] = dis[j][i] = x
            nei[i].append(j)
            nei[j].append(i)
        for i in range(n):
            dis[i][i] = 0
        
        # Dijkstra
        visited = set()
        for i in range(n):
            pool = [(0,i)]                                          # pool[j] = x:  d(i,j) = x    
            while pool:
                x,j = heappop(pool)                                 # x = d(i,j)
                if (i,j) not in visited and x <= distanceThreshold: # early stop, if distance exceeds threshold
                    visited.add((i,j))
                    for k in nei[j]:
                        dis[i][k] = min(dis[i][k], x+dis[j][k])     # dis(i,k) = min(dis(i,k), dis(i,j)+dis(j,k))
                        heappush(pool, (dis[i][k], k))
        cities = [[i,0] for i in range(n)]
        for i in range(n):
            for d in dis[i]:
                if d <= distanceThreshold:
                    cities[i][1] += 1
        result = sorted(cities,key=lambda x:[x[1],-x[0]])[0][0]
        return result

 

算法思路：图，单源最短路径，Dijkstra算法。

计算每一个点到其他可达点的最小路径，再与阈值进行比较。统计每一个城市为起点，满足阈值范围内的可到达的其他城市的数量，再使用双条件排序，得到最终结果。

参考：https://leetcode.com/problems/find-the-city-with-the-smallest-number-of-neighbors-at-a-threshold-distance/discuss/490614/Python-3-Dijkstra-Algorithm

24-29行，是我自己的代码，比原帖中的写的要容易理解一些。