如何列出无向图的所有派系? (并非所有最大派系,如Bron-Kerbosch算法)
答案 0 :(得分:0)
最优解是这样的,因为在完整的图中有2 ^ n个派系。使用递归函数考虑所有节点子集。对于每个子集,如果子集的节点之间存在所有边,则向计数器添加1 :(这几乎是C ++中的伪代码)
int clique_counter = 0;
int n; //number of nodes in graph
//i imagine nodes are numbered from 1 to n
void f(int x, vector <int> &v){ //x is the current node
if(x == n){
bool is_clique = true;
for(int i = 0; i < v.size(); i++){
for(int j = i + 1; j < v.size(); j++){
if(there is not an edge between node v[i] and v[j]) //it can't be clique
is_clique = false;
}
}
if(is_clique == true){
clique_counter++;
}
return;
}
//if x < n
f(x + 1, v);
v.push_back(x);
f(x + 1, v);
v.pop_back();
}
int main(){
vector <int> v;
f(1, v);
cout << clique_counter << endl;
return 0;
}