There are n
cities numbered from 0
to n-1
. Given the array edges
where edges[i] = [fromi, toi, weighti]
represents a bidirectional and weighted edge between cities fromi
and toi
, and given the integer distanceThreshold
.
Return the city with the smallest numberof cities that are reachable through some path and whose distance is at most distanceThreshold
, If there are multiple such cities, return the city with the greatest number.
Notice that the distance of a path connecting cities i and j is equal to the sum of the edges’ weights along that path.
Example 1:
Input: n = 4, edges = [[0,1,3],[1,2,1],[1,3,4],[2,3,1]], distanceThreshold = 4 Output: 3 Explanation: The figure above describes the graph. The neighboring cities at a distanceThreshold = 4 for each city are: City 0 -> [City 1, City 2] City 1 -> [City 0, City 2, City 3] City 2 -> [City 0, City 1, City 3] City 3 -> [City 1, City 2] Cities 0 and 3 have 2 neighboring cities at a distanceThreshold = 4, but we have to return city 3 since it has the greatest number.
Example 2:
Input: n = 5, edges = [[0,1,2],[0,4,8],[1,2,3],[1,4,2],[2,3,1],[3,4,1]], distanceThreshold = 2 Output: 0 Explanation: The figure above describes the graph. The neighboring cities at a distanceThreshold = 2 for each city are: City 0 -> [City 1] City 1 -> [City 0, City 4] City 2 -> [City 3, City 4] City 3 -> [City 2, City 4] City 4 -> [City 1, City 2, City 3] The city 0 has 1 neighboring city at a distanceThreshold = 2.
Constraints:
2 <= n <= 100
1 <= edges.length <= n * (n - 1) / 2
edges[i].length == 3
0 <= fromi < toi < n
1 <= weighti, distanceThreshold <= 10^4
- All pairs
(fromi, toi)
are distinct.
Solution1: Floyd-Warshall
All pair shortest path
Time complexity: O(n^3)
Space complexity: O(n^2)
C++
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// Author: Huahua class Solution { public: int findTheCity(int n, vector<vector<int>>& edges, int distanceThreshold) { vector<vector<int>> dp(n, vector<int>(n, INT_MAX / 2)); for (const auto& e : edges) dp[e[0]][e[1]] = dp[e[1]][e[0]] = e[2]; for (int k = 0; k < n; ++k) for (int u = 0; u < n; ++u) for (int v = 0; v < n; ++v) dp[u][v] = min(dp[u][v], dp[u][k] + dp[k][v]); int ans = -1; int min_nb = INT_MAX; for (int u = 0; u < n; ++u) { int nb = 0; for (int v = 0; v < n; ++v) if (v != u && dp[u][v] <= distanceThreshold) ++nb; if (nb <= min_nb) { min_nb = nb; ans = u; } } return ans; } }; |
Solution 2: Dijkstra’s Algorithm
Time complexity: O(V * ElogV) / worst O(n^3*logn), best O(n^2*logn)
Space complexity: O(V + E)
C++
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// Author: Huahua // Author: Huahua class Solution { public: int findTheCity(int n, vector<vector<int>>& edges, int t) { vector<vector<pair<int, int>>> g(n); for (const auto& e : edges) { g[e[0]].emplace_back(e[1], e[2]); g[e[1]].emplace_back(e[0], e[2]); } // Returns the number of nodes within t from s. auto dijkstra = [&](int s) { vector<int> dist(n, INT_MAX / 2); set<pair<int, int>> q; // <dist, node> vector<set<pair<int, int>>::const_iterator> its(n); dist[s] = 0; for (int i = 0; i < n; ++i) its[i] = q.emplace(dist[i], i).first; while (!q.empty()) { auto it = cbegin(q); int d = it->first; int u = it->second; q.erase(it); if (d > t) break; // pruning for (const auto& p : g[u]) { int v = p.first; int w = p.second; if (dist[v] <= d + w) continue; // Decrease key q.erase(its[v]); its[v] = q.emplace(dist[v] = d + w, v).first; } } return count_if(begin(dist), end(dist), [t](int d) { return d <= t; }); }; int ans = -1; int min_nb = INT_MAX; for (int i = 0; i < n; ++i) { int nb = dijkstra(i); if (nb <= min_nb) { min_nb = nb; ans = i; } } return ans; } }; |
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