Key cities--the plot of the cut point "AHA algorithm" code detailed

#include <iostream>using namespace Std;int n,m,e[9][9],root;int num[9],low[9],flag[9],index;void dfs (int cur,    int father) {int child=0;    index++;    Num[cur]=index;    Low[cur]=index;                          for (int i=1;i<=n;i++) {if (e[cur][i]==1) {//i is not a node that can be reached by the father if (num[i]==0) {///If it is Dad, it will not be 0 If not father, must be own ancestors//that is, you walked a ring//Because this is a deep search, of course, this ancestor is not necessarily the highest level of that, but, the total                than the father first access to the child++;                                              DFS (i,cur);//start looking for ancestors by their own son Low[cur]=min (Low[cur],low[i]);//Since cur can not go through dad to I, then, I can not through their father to reach the node, cur as can to                                              In this case, although cur than I first access, but it is very likely to find through I have visited a point already.                If you can find it, cur is not cutting it.                                                if (Cur!=root&&low[i]>=num[cur]) {///This is not found, because the initial value of low must be larger than NUM, so do not go through dad can not find the ancestors,                    The value of Low[i] is no smaller than num[cur].                Flag[cur]=1; } if (CUr==root&&child==2) {//child equals 2, it means the root node must be cut//because, only if his son has not been visited, will be his own son Simply put, at the beginning of the program, make sure that a node adjacent to it is the son, will immediately start deep search//Search                                        After the cable, other nodes adjacent to it, if accessed, will not enter the IF, so there will not be a new son//So the son number only two, the root node must be a cut point.                    The number of sons is three is also, but become three before, already became two when the time was confirmed, so write equals 2 no problem.                Flag[cur]=1;            }} else if (i!=father) {//not father, but ancestor Low[cur]=min (Low[cur],num[i]),//This sentence is the core AH!!! }}} return;}    int main () {int x, y;    cin>>n>>m;        for (int i=1;i<=n;i++) {for (int j=1;j<=n;j++) {e[i][j]=0;        }} for (int i=1;i<=m;i++) {cin>>x>>y;        E[x][y]=1;    E[y][x]=1;    } root=1;    DFS (1,root);    for (int i=1;i<n;i++) {if (flag[i]==1) {cout<<i<< "";    }}//cout<<low[6]<<endl;} 

　　

Key cities--the plot of the cut point "AHA algorithm" code detailed