 # CSES | SHORTEST ROUTES 2

Given n cities and m roads, find the shortest route distance for q queries (Each query consist of starting and ending city).
This problem is from the house of All Source Shortest Path. To solve this kind of problem, we have two approaches -
1. Use Single source shortest path algorithm for every vertex.
2. Use All Source shortest path explicitly.

Using Single Source Shortest Path Algorithm-
We can use Dijkstra's Algorithm for every vertex, so it will be like using Dijkstra's Algorithm n times.

The time complexity would be - O(n2 log n + n*m), using min heap.

But the problem with this method is, it fails for negative edges(which is not the case here, so this algorithm will work fine here).

Since we've already covered Dijkstra's Algorithm in previous article, so go with the second choice.

Using All Source Shortest Path Algorithm (Floyd-Warshall Algorithm)-

Floyd-Warshall Algorithm is a dynamic programming method to solve All Source Shortest Path problem.

Step 1 .

We create a distance matrix, say adj[n+1][n+1].

Before the ith pass, adj[j][k] stores the distance from vertex j to vertex k, which contains only vertices {1, 2, 3 ..., i-1} as internal vertices in the path.

For i = 0, adj[j][k] = distj,k if there exist a direct path between j and k, infinity otherwise.

Considering vertex i as bridge, we compare if we can relax the distance between two vertices j & k

The overall time complexity of the algorithm is O(n3).

Limitation:

Although graph may have negative edges, but there must not be a negative cycle.

C++ Implementation of Shortest Route II

#include<bits/stdc++.h>
using namespace std;
const int inf = 1e17;
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int n, m, q;
cin >> n >> m >> q;
vector<vector<int>> edges;
for(int i = 0; i < m; i++) {
int a, b, w;
cin >> a >> b >> w;
edges.push_back({a, b, w});
}
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= n; j++) {
if(i == j)
}
}
for(auto e : edges) {
int a = e, b = e, c = e;
}
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= n; j++) {
for(int k = 1; k <= n; k++) {
}
}
}
while(q--) {
int a, b;
cin >> a >> b;
cout << -1 << '\n';
else
cout << adj[a][b] << '\n';
}
return 0;
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