floyd-warshall Archives - Huahua's Tech Road

花花酱 LeetCode 2642. Design Graph With Shortest Path Calculator

zxi — Sat, 29 Apr 2023 19:16:35 +0000

There is a directed weighted graph that consists of n nodes numbered from 0 to n - 1. The edges of the graph are initially represented by the given array edges where edges[i] = [from_i, to_i, edgeCost_i] meaning that there is an edge from from_i to to_i with the cost edgeCost_i.

Implement the Graph class:

Graph(int n, int[][] edges) initializes the object with n nodes and the given edges.
addEdge(int[] edge) adds an edge to the list of edges where edge = [from, to, edgeCost]. It is guaranteed that there is no edge between the two nodes before adding this one.
int shortestPath(int node1, int node2) returns the minimum cost of a path from node1 to node2. If no path exists, return -1. The cost of a path is the sum of the costs of the edges in the path.

Example 1:

Input
["Graph", "shortestPath", "shortestPath", "addEdge", "shortestPath"]
[[4, [[0, 2, 5], [0, 1, 2], [1, 2, 1], [3, 0, 3]]], [3, 2], [0, 3], [[1, 3, 4]], [0, 3]]
Output
[null, 6, -1, null, 6]

Explanation

Graph g = new Graph(4, [[0, 2, 5], [0, 1, 2], [1, 2, 1], [3, 0, 3]]); g.shortestPath(3, 2); // return 6. The shortest path from 3 to 2 in the first diagram above is 3 -> 0 -> 1 -> 2 with a total cost of 3 + 2 + 1 = 6.
g.shortestPath(0, 3); // return -1. There is no path from 0 to 3. 
g.addEdge([1, 3, 4]); // We add an edge from node 1 to node 3, and we get the second diagram above. 
g.shortestPath(0, 3); // return 6. The shortest path from 0 to 3 now is 0 -> 1 -> 3 with a total cost of 2 + 4 = 6.

Constraints:

1 <= n <= 100
0 <= edges.length <= n * (n - 1)
edges[i].length == edge.length == 3
0 <= from_i, to_i, from, to, node1, node2 <= n - 1
1 <= edgeCost_i, edgeCost <= 10⁶
There are no repeated edges and no self-loops in the graph at any point.
At most 100 calls will be made for addEdge.
At most 100 calls will be made for shortestPath.

Solution 1: Floyd-Washall

Time complexity:
Init O(n³)
addEdge O(n²)
shortestPath O(1)

Space complexity: O(1)

C++

// Author: Huahua
class Graph {
public:
  Graph(int n, vector>& edges):
  n_(n),
  d_(n, vector(n, 1e18)) {
    for (int i = 0; i < n; ++i)
      d_[i][i] = 0;
    for (const vector& e : edges)
      d_[e[0]][e[1]] = e[2];    
    for (int k = 0; k < n; ++k)
      for (int i = 0; i < n; ++i)
        for (int j = 0; j < n; ++j)
          d_[i][j] = min(d_[i][j], d_[i][k] + d_[k][j]);
  }
  
  void addEdge(vector edge) {    
    for (int i = 0; i < n_; ++i)
      for (int j = 0; j < n_; ++j)
        d_[i][j] = min(d_[i][j], d_[i][edge[0]] + edge[2] + d_[edge[1]][j]);
  }
  
  int shortestPath(int node1, int node2) {
    return d_[node1][node2] >= 1e18 ? -1 : d_[node1][node2];
  }
private:
  int n_;
  vector> d_;
};

Solution 2: Dijkstra

Time complexity:
Init: O(|E|) ~ O(n²)
AddEdge: O(1)
ShortestPath: O(|V|*log(|E|)) ~ O(n*logn)

Space complexity: O(E|) ~ O(n²)

C++

// Author: Huahua
class Graph {
public:
    Graph(int n, vector>& edges): n_(n), g_(n) {
      for (const auto& e : edges)
        g_[e[0]].emplace_back(e[1], e[2]);
    }
    
    void addEdge(vector edge) {
      g_[edge[0]].emplace_back(edge[1], edge[2]);
    }
    
    int shortestPath(int node1, int node2) {
      vector dist(n_, INT_MAX);
      dist[node1] = 0;
      priority_queue, vector>, greater<>> q;
      q.emplace(0, node1);
      while (!q.empty()) {
        const auto [d, u] = q.top(); q.pop();        
        if (u == node2) return d;
        for (const auto& [v, c] : g_[u])
          if (dist[u] + c < dist[v]) {
            dist[v] = dist[u] + c;
            q.emplace(dist[v], v);
          }
      }
      return -1;
    }
private:
  int n_;
  vector>> g_;
};

The post 花花酱 LeetCode 2642. Design Graph With Shortest Path Calculator appeared first on Huahua's Tech Road.

花花酱 LeetCode 1334. Find the City With the Smallest Number of Neighbors at a Threshold Distance

zxi — Mon, 27 Jan 2020 06:58:54 +0000

There are n cities numbered from 0 to n-1. Given the array edges where edges[i] = [from_i, to_i, weight_i] represents a bidirectional and weighted edge between cities from_i and to_i, 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 <= from_i < to_i < n
1 <= weight_i, distanceThreshold <= 10^4
All pairs (from_i, to_i) are distinct.

Solution1: Floyd-Warshall

All pair shortest path

Time complexity: O(n^3)
Space complexity: O(n^2)

C++

// Author: Huahua
class Solution {
public:
  int findTheCity(int n, vector>& edges, int distanceThreshold) {
    vector> dp(n, vector(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++

// Author: Huahua
// Author: Huahua
class Solution {
public:
  int findTheCity(int n, vector>& edges, int t) {
    vector>> 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 dist(n, INT_MAX / 2);
      set> q; // 
      vector>::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;
  }
};

The post 花花酱 LeetCode 1334. Find the City With the Smallest Number of Neighbors at a Threshold Distance appeared first on Huahua's Tech Road.