Module: Dijkstra's algorithm


Problem

8/14

Dijkstra's algorithm in O(M logN) c set: Start (C++)

Theory Click to read/hide

Since the asymptotic behavior of the naive implementation of Dijkstra's algorithm is: \(O(n^2 + m)\), then as the number of vertices increases, the speed of work becomes unsatisfactory.
 Various data structures can be used for improvement: Fibonacci heaps, set sets, or a priority queue priority_queue. 
Consider an example with set, as a result, the final asymptotics is: \(O(n log (m))\), details.

Problem

You are given a directed weighted graph. Find the shortest distance from one given vertex to another.
 
Input
The first line contains three numbers: N, M, S and F (1≤ N≤ 100, 1≤ S, F≤ N), where N – number of graph vertices, M – number of ribs,  S– initial vertex, and F – final. In the next N lines, enter N numbers each, not exceeding 100, – graph adjacency matrix, where -1 means no edge between vertices, and any non-negative number – the presence of an edge of given weight. Zeros are written on the main diagonal of the matrix.
 
Output
It is required to display the desired distance or -1 if there is no path between the specified vertices.

Examples
# Input Output
1 4 4 3 4
3 1 3
1 2 3
2 4 3
3 4 10 		9

time 1000 ms
memory 256 Mb
Rules for program design and list of errors in automatic problem checking

Teacher commentary

Insert the missing code snippets
C++
Write the program below 
#include <iostream>
#include <set>
#include <vector>
using namespace std;

	const int INF = 1000000000;

	int main() {
		int n, m ,s, f;
		cin >> n>>m>>s>>f;
		
		vector < vector < pair<int, int> > > g(n+1);
		// чтение графа
    
		vector<int> d(n+1, INF);
		d[s] = 0;
		set < pair<int, int> > q;
		q.insert(make_pair(d[s], s));
		while (!q.empty()) {
			int v = q.begin()->second;
			q.erase(q.begin());

			for (size_t j = 0; j < g[v].size(); ++j) {
				int to = g[v][j].first,
					len = g[v][j].second;
				if (d[v] + len < d[to]) {
					q.erase(make_pair(d[to], to));
					d[to] = d[v] + len;
					q.insert(make_pair(d[to], to));
				}
			}
		}

		if (d[f] == 10000000)
			cout << "-1";
		else
			cout << d[f];
	}

     

Program check result

To check the solution of the problem, you need to register or log in!