C++ 程式檢查無向圖的連通性使用 BFS
想要檢查圖的連線性,我們將嘗試使用任意遍歷演算法遍歷所有節點。遍歷完成後,如果仍有未訪問的節點,則說明圖不是連通的。
對於無向圖,我們將選擇一個節點並從中進行遍歷。
本例中,遍歷演算法是遞迴 BFS 遍歷。
輸入 − 圖的鄰接矩陣
0 | 1 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 1 | 1 |
0 | 1 | 1 | 0 | 1 |
0 | 0 | 1 | 1 | 0 |
輸出 − 圖已連線。
演算法
traverse(s, visited)
輸入 − 開始節點 s 和 visited 節點,用於標記哪個節點已被訪問。
輸出 − 遍歷所有連線的頂點。
Begin mark s as visited insert s into a queue Q until the Q is not empty, do u = node that is taken out from the queue for each node v of the graph, do if the u and v are connected, then if u is not visited, then mark u as visited insert u into the queue Q. done done End
isConnected(graph)
輸入 − 圖。
輸出 − 如果圖連通,則返回 True。
Begin define visited array for all vertices u in the graph, do make all nodes unvisited traverse(u, visited) if any unvisited node is still remaining, then return false done return true End
示例程式碼 (C++)
#include<iostream> #include<queue> #define NODE 5 using namespace std; int graph[NODE][NODE] = { {0, 1, 1, 0, 0}, {1, 0, 1, 1, 0}, {1, 1, 0, 1, 1}, {0, 1, 1, 0, 1}, {0, 0, 1, 1, 0}}; void traverse(int s, bool visited[]) { visited[s] = true; //mark v as visited queue<int> que; que.push(s);//insert s into queue while(!que.empty()) { int u = que.front(); //delete from queue and print que.pop(); for(int i = 0; i < NODE; i++) { if(graph[i][u]) { //when the node is non-visited if(!visited[i]) { visited[i] = true; que.push(i); } } } } } bool isConnected() { bool *vis = new bool[NODE]; //for all vertex u as start point, check whether all nodes are visible or not for(int u; u < NODE; u++) { for(int i = 0; i < NODE; i++) vis[i] = false; //initialize as no node is visited traverse(u, vis); for(int i = 0; i < NODE; i++) { if(!vis[i]) //if there is a node, not visited by traversal, graph is not connected return false; } } return true; } int main() { if(isConnected()) cout << "The Graph is connected."; else cout << "The Graph is not connected."; }
輸出
The Graph is connected.
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