点双连通分量

在Tarjan算法的过程中维护一个栈,并按如下方法维护其中的元素
1:当一个节点第一次被访问时,入栈。
2:当割点判定法则中dfn[x]<=Low[y]成立时
无论X是否为根,都要
1:从栈顶不断顶出节点,直到节点Y被弹出
2:刚才弹出的所有节点与节点X一起构成一个V-DCC
注意节点X还在栈中

#include
#include
#include
#include
#include
#include
using namespace std;
const int N = 20010, M = 200010;
int head[N], ver[M], Next[M];
int dfn[N], low[N], stack[N], new_id[N], c[N], belong[M];
int d[N], dist[N], f[N][16];
int n, m, t, tot, num, root, top, cnt, tc;
bool cut[N];
vector dcc[N];
int hc[N], vc[M], nc[M];
queue q;

void add(int x, int y) 
{
	ver[++tot] = y, Next[tot] = head[x], head[x] = tot;
}

void add_c(int x, int y) 
{
	vc[++tc] = y, nc[tc] = hc[x], hc[x] = tc;
}

void tarjan(int x) 
{
	dfn[x] = low[x] = ++num;
	stack[++top] = x;
	if (x == root && head[x] == 0) 
	{ 
		dcc[++cnt].push_back(x);
		return;
	}

	int flag = 0;
	for (int i = head[x]; i; i = Next[i]) 
	{
		int y = ver[i];
		if (!dfn[y]) 
		{
			tarjan(y);
			low[x] = min(low[x], low[y]);
			if (low[y] >= dfn[x]) 
			//x-->y,发现low[y]>=dfn[x],则X是一个割点 
			{
				flag++;
				if (x != root || flag > 1) //如果X不为根,或者X为根,但有两个子树时 
				      cut[x] = true;
				cnt++;
				int z;
			
				do 
				//将栈中的元素不断弹出来,直到Y这个结点 
				{
					z = stack[top--];
					dcc[cnt].push_back(z);
				
				} 
				while (z != y);
				dcc[cnt].push_back(x);//将X这个点也加入点双中,但X仍在栈中 
			
			}
		}
		else low[x] = min(low[x], dfn[y]);
	}
}


int main() 
{
	    cin>>n>>m;
		memset(head, 0, sizeof(head));
		memset(hc, 0, sizeof(hc));
		memset(dfn, 0, sizeof(dfn));
		memset(d, 0, sizeof(d));
		memset(cut, 0, sizeof(cut));
		memset(c, 0, sizeof(c));
		for (int i = 1; i <= n; i++) dcc[i].clear();
		tot = 1; num = cnt = top = 0;
		for (int i = 1; i <= m; i++) 
		{
			int x, y;
			scanf("%d%d", &x, &y);
			add(x, y), add(y, x);
		}
		for (int i = 1; i <= n; i++)
			if (!dfn[i]) root = i, tarjan(i);
		for (int i=1;i<=cnt;i++)
		{
			cout<<"e-dcc   "< 
 

  运行结果

e-dcc 1
4 3 2
e-dcc 2
2 1

 

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