Sicily shortest path in unweighted graph, sicilyunweighted

Sicily shortest path in unweighted graph, sicilyunweighted

Topic introduction:

Enter an undirected graph, specify a vertex s to start bfs traversal, and obtain the shortest distance from s to each vertex in the graph.

If the path from s to t does not exist, the distance from s to t is-1. Input

The first line of the input contains two integers n and m. n indicates the number of vertices of the graph, and m indicates the number of edges. 1 <= n <= 1000,0 <= m <= 10000.

In the following m rows, each row is a number pair v y, indicating that an edge (v, y) exists ). Vertex number starts from 1. Output

Note s = 1, and output in one row: the shortest distance from vertex 1 to vertex s, the shortest distance from vertex 2 to vertex s,..., the shortest distance from vertex n to vertex s.

Add a space after each output, including the last one. Sample Input

5 31 21 32 4

Sample Output

0 1 1 2 -1 

Ideas:

Use the breadth search function to mark the number of layers and calculate the distance in sequence.

The Code is as follows:

 1 #include <iostream> 2 #include <queue> 3 using namespace std; 4  5 bool path[1001][1001]; 6 int shortest[1001]; 7  8 int main() { 9     int n, m;10     cin >> n >> m;11     12     for (int i = 1; i <= m; i++) {13         int node1, node2;14         cin >> node1 >> node2;15         path[node1][node2] = true;16         path[node2][node1] = true;17     }18     19     for (int i = 1; i <= n; i++)20         i == 1 ? shortest[i] = 0 : shortest[i] = -1;21     22     int distance = 0;23     queue<int> store;24     store.push(1);25     while (!store.empty()) {26         int size = store.size();27         distance++;28         while (size--) {29             for (int i = 1; i <= n; i++) {30                 if (path[store.front()][i] && shortest[i] == -1) {31                     shortest[i] = distance;32                     store.push(i);33                 }34             }35             store.pop();36         }37     }38     39     for (int i = 1; i <= n; i++)40         cout << shortest[i] << " ";41     cout << endl;42     43     return 0;44 }