边双连通分量

找出所有的桥出来，然后去掉桥即可

#include
#include
using namespace std;
const int MAXN=1e5,MAXM=1e6;
struct Edge{
	int from,to,nxt;
}e[MAXM];
int head[MAXN],edgeCnt=1;
void addEdge(int u,int v)
{
	e[++edgeCnt].from=u;
	e[edgeCnt].to=v;
	e[edgeCnt].nxt=head[u];
	head[u]=edgeCnt;
}
int dfn[MAXN],low[MAXN],dfnCnt=0;
bool bridge[MAXM];
void tarjan(int x,int in_edge)
{
	dfn[x]=low[x]=++dfnCnt;
	for(int i=head[x];i;i=e[i].nxt)
	{
		int nowV=e[i].to;
		if(!dfn[nowV])
		{
			tarjan(nowV,i);
			if(low[nowV]>dfn[x])
			{
				bridge[i]=bridge[i^1]=1;
			}
			low[x]=min(low[x],low[nowV]);
		}
		else 
		    if(i!=(in_edge^1))
		     {
			     low[x]=min(low[x],dfn[nowV]);
		     }
	}
}
int inDcc[MAXN];
void dfs(int x,int nowDcc)
{
	inDcc[x]=nowDcc;//将X点归入nowdcc这个边双中 
	for(int i=head[x];i;i=e[i].nxt)
	{
		int nowV=e[i].to;
		if(inDcc[nowV]||bridge[i])
		//如果已归到某个边双，或第I条边是桥的话 
		    continue;
		dfs(nowV,nowDcc);
	}
}
int main(){
	int n,m;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		addEdge(u,v);
		addEdge(v,u);
	}
	for(int i=1;i<=n;i++) //找出所有的桥出来 
		if(!dfn[i])
		    tarjan(i,0);
	int nowDcc=0;
	for(int i=1;i<=n;i++) //如果i点还没有归到某个边双 
		if(!inDcc[i])
		   dfs(i,++nowDcc);
	printf("%d\n",nowDcc);
	return 0;
}