图的连通性问题模板

图的点双连通分量

连通分量中无割点,即除去一个点及其边,图仍然连通

void tarjan(int u,int fa0){//点双
	dfn[u] = low[u] = ++ cnt;
	for (int i = 0;i < G[u].size();i ++){
		int v = G[u][i];
		if (!dfn[v]){
			s[++ siz] = v;
			tarjan(v,u);
			if (low[v] >= dfn[u]){
				num ++;
				while (1){
					int t1;
					t1 = s[siz];
					fa[t1].push_back(num);
					tot[num] ++;
					siz --;
					if (t1 == u)
						break;
				}
			}
			low[u] = min(low[u],low[v]);
		}
		else if (v != fa0 && dfn[v] < dfn[u]){
			s[++ siz] = v;
			low[u] = min(low[u],dfn[v]);
		}
	}
}

图的边双连通分量

图的连通分量中无割边,即去除一条边,图仍然连通

void tarjan(int u,int fa0){
	dfn[u] = low[u] = ++cnt;
	s[++ siz] = u;
	for (int i = 0;i < G[u].size();i ++){
		int v = G[u][i];
		if (!dfn[v]){
			tarjan(v,u);
			low[u] = min(low[u],low[v]);
		}
		else if (v != fa0)
			low[u] = min(low[u],dfn[v]);
	}
	if (low[u] == dfn[u]){
		num ++;
		fa[u] = num;
		while (s[siz] != u){
			fa[s[siz]] = num;
			siz --;
		}
		siz --;
	}
}

强连通分量

有向图的连通分量中每两个点都可以互相到达

void tarjan(int u){
	dfn[u] = low[u] = ++ cnt;
	s[++ siz] = u;
	instuck[u] = 1;
	for (int i = 0;i < G[u].size();i ++){
		int v = G[u][i];
		if (!dfn[v]){
			tarjan(v);
			low[u] = min(low[u],low[v]);
		}
		if (instuck[v])
			low[u] = min(low[u],low[v]);
	}
	if (low[u] == dfn[u]){
		num ++;
		instuck[u] = 0;
		while (s[siz] != u){
			instuck[s[siz]] = 0;
			siz --;
		}
		siz --;
	}
}
原文地址:https://www.cnblogs.com/lover-fucker/p/13566653.html