三种方法求图中连通分量的个数(BFS、DFS、并查集)

三种方法求图中连通分量的个数(BFS、DFS、并查集)

1. 连通分量是什么

无向图G的极大连通子图称为G的连通分量( Connected Component)。任何连通图的连通分量只有一个,即是其自身,非连通的无向图有多个连通分量。

2. 案例

2.1.图极其数据结构初始化

2.2.求连通分量的方法

从每个顶点出发,判断是否有连通分量

BFS[BFS](https://blog.csdn.net/qq_44423388/article/details/127591933?spm=1001.2014.3001.5501)

DFS[DFS](https://blog.csdn.net/qq_44423388/article/details/127583096?spm=1001.2014.3001.5501)

并查集(本篇主讲,实现步骤见下)

2.3 具体实现

/*

测试用例:

1 2

1 4

2 4

*/

#include

#include

#include

#include

using namespace std;

/*

如果节点是相互连通的(从一个节点可以到达另一个节点),那么他们在同一棵树里,或者说在同一个集合里,或者说他们的祖先是相同的。

*/

//并查集的数据结构

class UnionFind {

private:

// 记录每一个节点的父节点father<当前节点下标,父节点下标>

unordered_map father;

// 记录集合数量

int num_of_sets = 0;

public:

//找节点x的父节点

int find(int x)

{

int root = x;

while (father[root] != -1)

{

root = father[root];

}

//优化的点:如果我们树很深,那么每次查询的效率都会非常低。这一步把树的深度固定为二。

while (x != root)

{

int original_father = father[x];

father[x] = root;

x = original_father;

}

return root;

}

bool is_connected(int x, int y)

{

return find(x) == find(y);

}

//将连通的两个节点合并为同一个祖先,同时并查集的数目--

void merge(int x, int y)

{

int root_x = find(x);

int root_y = find(y);

if (root_x != root_y)

{

father[root_y] = root_x;

num_of_sets--;

}

}

//将新节点添加到并查集中

void add(int x)

{

if (!father.count(x))

{

father[x] = -1;

num_of_sets++;

}

}

//返回并查集个数

int get_num_of_sets()

{

auto it = father.begin();

while (it != father.end())

{

cout << it->first<<" ->"<second << endl;

it++;

}

return num_of_sets;

}

};

class Connectedcomponent:protected UnionFind

{

private:

int vertice = 0;//顶点数

int edge = 0;//边数

vector> e;

//因为dfs和bfs都会对其进行改变,所有设置两个book

vector book;//判断顶点j是否扩展过

vector book1;//判断顶点j是否扩展过

queue qu;

//DFS求连通分量个数

void DFS_Alg(int current, int sum)//current当前所在的节点编号

{

sum++;

if (sum == vertice)//所有的节点均已被访问

{

cout << current << endl;

return;

}

else

{

cout << current << " ->";

}

for (int k = 1; k <= vertice; k++)

{

if (e[current][k] != 0 && book[k] == 0)

{

book[k] = 1;

DFS_Alg(k, sum);

}

}

}

public:

Connectedcomponent(int x, int y) :vertice(x), edge(y)

{

//图的初始化从下标1开始

e.resize(vertice + 1);//初始化二维数组的行

for (int i = 0; i <= vertice; i++)

{

e[i].resize(vertice + 1,0);//初始化二维数组的列

}

book.resize(vertice + 1);

book1.resize(vertice + 1);

}

//图的初始化

void Init_tu()

{

for (int i = 0; i <= vertice; i++)

{

for (int j = 0; j <= vertice; j++)

{

if (i == 0 || j == 0)

{

e[i][j] = 0;

}

if (i == j)

{

e[i][j] = 0;

}

else

{

e[i][j] = INT_MAX;

}

}

}

}

//读入图的边,并且根据边的信息初始化数组dis,数组book

void GetEdgeInfo()

{

cout << "输入边的信息(节点1,节点2):" << endl;

int e1 = 0, e2 = 0, weigth = 0;

for (int i = 1; i <= edge; i++)//无向图

{

cin >> e1 >> e2;

e[e1][e2] = 1;

e[e2][e1] = 1;

}

}

//打印

void Print()

{

for (int i = 1; i <= vertice; i++)

{

for (int j = 1; j <= vertice; j++)

{

cout << e[i][j] << " ";

}

cout << endl;

}

cout << endl;

}

int DFS_Num()

{

int num = 0;

for (int i = 1; i <= vertice; i++)

{

if (book[i] == false)

{

DFS_Alg(i,0);

cout <<"end" <

num++;

}

}

return num;

}

//BFS求连通分量个数

int BFS_Num()

{

int num = 0;

for (int i = 1; i <= vertice; i++)//遍历每个节点,查看是否从该节点出发是否有连通分量

{

if (book1[i] == false)

{

qu.push(i);

while (!qu.empty())

{

int v = qu.front();

qu.pop();

book1[v] = true;

cout << v << "->";

for (int i = 1; i <= vertice; i++)//循坏找节点v的相邻节点

{

if (e[v][i] != 0 && book1[i] == false)

{

qu.push(i);

book1[i] = true;

}

}

}

num++;

}

cout << "end" << endl;

}

return num;

}

//并查集求连通分量的个数

/*

每个节点会记录它的父节点。

*/

int UnionFindSet()

{

UnionFind uf;

for (int i = 1; i <= vertice; i++)

{

uf.add(i);

for (int j = 1; j < i; j++)

{

if (e[i][j] == 1)

{

uf.merge(i, j);

}

}

}

return uf.get_num_of_sets();

}

};

int main()

{

int num1 = 0, num2 = 0,num3 = 0;

Connectedcomponent Conn(5, 3);

Conn.GetEdgeInfo();

cout << "初始信息:" << endl;

Conn.Print();

cout << "DFS:::" << endl;

num1 = Conn.DFS_Num();

cout << "BFS:::" << endl;

num2 = Conn.BFS_Num();

cout << "Union Find Set:::" << endl;

num3 = Conn.UnionFindSet();

cout << num1 << " " << num2 <<" "<

return 0;

}

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