bfs廣度優先搜索算法

What you will learn?

您將學到什么？

How to implement Breath first search of a graph?

如何實現圖的呼吸優先搜索？

Breadth First Search is a level-wise vertex traversal process. Like a tree all the graphs have vertex but graphs have cycle so in searching to avoid the coming of the same vertex we prefer BFS

Algorithm:

To implement the BFS we use queue and array data structure.

There are two cases in the algorithm:

Whenever we visit a vertex we mark it visited and push its adjacent non-visited vertices into the queue and pop the current vertex from the queue.

If all the adjacent vertices are visited then only pop the current vertex from the queue.

Consider this graph,

According to our algorithm, the traversal continues like,

Hence all the vertices are visited then only pop operation is performed and queue will be empty finally.

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C++ Implementation:

#include <bits/stdc++.h> using namespace std;// Make a pair between vertex x and vertex y void addedge(list<int> *ls,int x,int y){ls[x].push_back(y);ls[y].push_back(x);return; } //Breath First Search of a Graph void BFS(list<int>*ls,int num,int x){bool *visit= new bool[num];for(int i=0;i<num;i++){visit[i]=false;}queue<int> q;q.push(x);while(!q.empty()){int s=q.front();q.pop();if(!visit[s]){visit[s]=true;cout<<s<<" ";list<int>::iterator it;for(it=ls[s].begin();it!=ls[s].end();it++){q.push(*it);}}} }// Print the Adjacency List void print(list<int> *ls,int num){list<int>::iterator it;for(int i=0;i<6;i++){cout<<i<<"-->";for(it=ls[i].begin();it!=ls[i].end();it++){cout<<*it<<"-->";}cout<<endl;} }int main(){int num=6;cout<<"Enter the no. of vertices : 6\n";list<int> *ls=new list<int>[num];addedge(ls,0,2);addedge(ls,2,3);addedge(ls,3,4);addedge(ls,4,5);addedge(ls,2,5);addedge(ls,1,4);addedge(ls,3,0);cout<<"Print of adjacency list:"<<endl;print(ls,6);cout<<"BFS"<<endl;BFS(ls,6,0);return 0; }

Output

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Comments and Discussions

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廣度優先搜索是一個逐層的頂點遍歷過程。 像一棵樹一樣，所有圖都具有頂點，但是圖卻具有循環，因此為了避免出現相同的頂點，我們選擇了BFS

算法：

為了實現BFS，我們使用隊列和數組數據結構。

該算法有兩種情況：

每當我們訪問頂點時，都會將其標記為已訪問，并將其相鄰的未訪問頂點推入隊列，然后從隊列中彈出當前頂點。

如果訪問了所有相鄰的頂點，則僅從隊列中彈出當前頂點。

考慮一下這張圖，

根據我們的算法，遍歷繼續像

因此，所有頂點都將被訪問，然后僅執行彈出操作，并且隊列最終將為空。

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C ++實現：

# include < bits/stdc++.h > using namespace std ;// Make a pair between vertex x and vertex y void addedge ( list < int > * ls , int x , int y ) {ls [ x ] . push_back ( y ) ;ls [ y ] . push_back ( x ) ;return ; } //Breath First Search of a Graph void BFS ( list < int > * ls , int num , int x ) {bool * visit = new bool [ num ] ;for ( int i = 0 ; i < num ; i + + ) {visit [ i ] = false ;}queue < int > q ;q . push ( x ) ;while ( ! q . empty ( ) ) {int s = q . front ( ) ;q . pop ( ) ;if ( ! visit [ s ] ) {visit [ s ] = true ;cout < < s < < " " ;list < int > :: iterator it ;for ( it = ls [ s ] . begin ( ) ; it ! = ls [ s ] . end ( ) ; it + + ) {q . push ( * it ) ;}}} }// Print the Adjacency List void print ( list < int > * ls , int num ) {list < int > :: iterator it ;for ( int i = 0 ; i < 6 ; i + + ) {cout < < i < < " --> " ;for ( it = ls [ i ] . begin ( ) ; it ! = ls [ i ] . end ( ) ; it + + ) {cout < < * it < < " --> " ;}cout < < endl ;} }int main ( ) {int num = 6 ;cout < < " Enter the no. of vertices : 6 \n " ;list < int > * ls = new list < int > [ num ] ;addedge ( ls , 0 , 2 ) ;addedge ( ls , 2 , 3 ) ;addedge ( ls , 3 , 4 ) ;addedge ( ls , 4 , 5 ) ;addedge ( ls , 2 , 5 ) ;addedge ( ls , 1 , 4 ) ;addedge ( ls , 3 , 0 ) ;cout < < " Print of adjacency list: " < < endl ;print ( ls , 6 ) ;cout < < " BFS " < < endl ;BFS ( ls , 6 , 0 ) ;return 0 ; }

輸出量

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• 檢查字謎(檢查兩個字符串是否是字謎)

• 相對排序算法

• 查找給定總和的子數組

• 在給定總和K的二叉樹中找到級別

• 檢查二叉樹是否為BST(二叉搜索樹)

• 1 [0] 1個樣式計數

• 大寫一行中每個單詞的第一個和最后一個字母

• 打印二叉樹的垂直和

• 打印二叉樹的邊界和

• 反轉單個鏈表

• 解決主要算法問題的貪婪策略

• 工作排序問題

• 根到葉的路徑總和

• 矩陣中的出口點

• 在鏈表中查找循環長度

• 一流的禮帽

• 打印所有沒有兄弟的節點

• 轉換為求和樹

• 最短的源到目標路徑

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翻譯自: https://www.includehelp.com/data-structure-tutorial/breadth-first-search-bfs-of-a-graph.aspx

bfs廣度優先搜索算法

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