通过邻接矩阵java的简单连接图

Question

I am trying to make a program which decides if a graph is simple connected or not, using the adjacency matrix. I managed to make the code to tell me that all the nodes have a link to something, but that doesn't guarantee me that there is a way between the first and the last node(definition of simple connected graph). I am a beginner so I don't know if this kind of question is ok to be posted here or not. Any idea how could I do this ? Here's my code so far. Any suggestion is appreciated.

package lab41;

import java.util.Scanner;

public class Graf {

    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        System.out.println("Introduceti nr de noduri:");
        int n = s.nextInt();
        int[][] a = new int[n + 1][n + 1];
        int i, j, k;
        System.out.println("Introduceti matricea adiacenta:");

        for (i = 1; i <= n; i++)
            for (j = 1; j <= n; j++) {
                System.out.println("a[" + i + "][" + j + "]=");
                a[i][j] = s.nextInt();
            }

        int x = 0;
        j = 1;
        i = 1;
        do {
            if (a[i][j] == 1) {
                i++;
                j = 1;
                x++;
            } else
                j++;

        }
        while (i < n && j <= n);

        if (x == n - 1)
            System.out.println("Graful este conex");
        else
            System.out.println("Graful nu este conex");
    }


}

Answer 1

To find if a Graph is connected you can perform a DFS & BFS starting from any node and if all the nodes are visited this means that the graph is connected.

Algorithm:

1. Assign a list, visitedNodes = {} , and a queue={}
2. pick any node , say Node 0.
3. Start BFS from Node 0. queue = {Node 0}
4. Pop the head of the queue , say the Node is x. If node x is in visitedNodes go to step 3. Else put x in visitedNodes list.
5. Visit all neighbors of Node x, and place them in a Queue
6. Continue steps 3 to 5 till queue is empty