求无向图连通图的割点

无向连通图的割点，即为除去该点，不再连通。

求割点数量， 参考代码(待续)：

#include<stdio.h>

#define MAX_LEN 100

int network[MAX_LEN + 2][MAX_LEN + 2];

int low[MAX_LEN+2];
int dnf[MAX_LEN+2];
bool visited[MAX_LEN+2];
int tmpDnf;
int critcalCnt;
int N;
int son;
int subNet[MAX_LEN+2];

void init(){
	int i, j;
	critcalCnt = 0;
	tmpDnf = 1;
	son = 0;
	for (i = 1; i <= MAX_LEN; i++) {
		low[i] = dnf[i] = subNet[i] = 0;
		visited[i] = false;
		for(j = 1; j <= MAX_LEN; j++) {
			network[i][j] = 0;
		}
	}
}

int getMin(int num1, int num2) {
	return num1 > num2 ? num2: num1;
}

void dfs(int u) {
	int v;
	for(v = 1; v <= N; v++) {
		if(network[u][v] == 1) {
			if(visited[v] == false) {
				tmpDnf++;
				low[v] = dnf[v] = tmpDnf;
				visited[v] = true;
				dfs(v);
				low[u] = getMin(low[u], low[v]);
				if(low[v] >= dnf[u]) {
					if(u == 1){
						son++; // 根节点需要至少两个子女
					} else {
						subNet[u]++;
					}
				}
			} else {
				low[u] = getMin(low[u], dnf[v]);
			}
		}
	}
}

void countCritical(){
	int i;
	for(i = 1; i <=N; i++) {
		if(subNet[i] > 0)   // 分成子网络个数 为 subNet[i] +1 ，包括它的祖先
			critcalCnt++;
	}
}

int main(){
	int a, b, c;

	//freopen("input.txt", "r", stdin);

	while(scanf("%d",&N),N){
		init();
		while(scanf("%d",&a),a){
			while(1){
				c = getchar();
				if(c == '\n')break;
				scanf("%d",&b);
				network[a][b] = 1;
				network[b][a] = 1;
			}
		}
	    low[1] = dnf[1] = 1;
		visited[1] = true;
		dfs(1);
		if(son > 1) // 根节点需要至少两个子女
			subNet[1] = son -1;
		countCritical();
		printf("%d\n", critcalCnt);
	}
	return 0;
}

参考链接：

http://blog.youkuaiyun.com/yangkunpengd/article/details/51298144

https://www.cnblogs.com/llhthinker/p/4954082.html#autoid-0-0-0