割点

和强连通分量十分相似
分为树枝边、前向边和后向边
注意！

if(x!=r&&low[to]>=dfn[x]) cut[x]=1;

这句判断只能在树枝边出现
否则会因为前向边的存在而出错

代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
int n,m;
const int N=1e6+100;
struct node{int to,nxt;
}p[2*N];
int fi[N],cnt=-1;
void addline(int x,int y){p[++cnt]=(node){y,fi[x]};fi[x]=cnt;
}int r;
int dfn[N],t,low[N],cut[N];
void tj(int x){dfn[x]=low[x]=++t;int num=0;for(int i=fi[x];~i;i=p[i].nxt){int to=p[i].to;if(!dfn[to]){
//		printf("x=%d to=%d\n",x,to);tj(to);low[x]=min(low[x],low[to]);if(x!=r&&low[to]>=dfn[x]) cut[x]=1;num++;}low[x]=min(low[x],dfn[to]);}if(x==r&&num>1) cut[x]=1;return;
}
int main(){memset(fi,-1,sizeof(fi));scanf("%d%d",&n,&m);int a,b,c;for(int i=1;i<=m;i++){scanf("%d%d",&a,&b);addline(a,b);addline(b,a);	}for(int i=1;i<=n;i++){r=i;if(!dfn[i]) tj(i);}int ans=0;for(int i=1;i<=n;i++){if(cut[i]){ans++;}}printf("%d\n",ans);for(int i=1;i<=n;i++){if(cut[i]){printf("%d ",i);}}return 0;
}

桥

也是类似的tarjan求法
但是需要注意:

1. 树枝边需要直接跳过
2. 判断的条件是$lo{w}_{to}<df{n}_x$（不带等！！）

仔细想想都是有道理的
2.是因为子树如果有连回x的边那么当前边也不是割边
1.是为了2服务的

void tarjan(int x){dfn[x]=low[x]=++tim;for(int i=fi[x];~i;i=p[i].nxt){if(i==(from[x]^1)) continue;int to=p[i].to;//printf("  x=%d to=%d\n",x,to);if(!dfn[to]){from[to]=i;tarjan(to);low[x]=min(low[x],low[to]);if(low[to]>dfn[x]) ans++;}low[x]=min(low[x],dfn[to]);}
}

点双连通分量

定义：点连通度>1的极大子图

考虑建一个栈
把所有的前向边和树枝边压入栈中
每次通过$df{n}_x<=lo{w}_{to}$，判断x是隔点时，就把栈里$(u,v)$及上面的边全部弹出，他们相连的点组成一个点双连通分量
找分量的好方法：dfs！！

代码

洛谷P3225

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N=2e3+100;
const double eps=1e-6;
inline ll read(){ll x=0,f=1;char c=getchar();while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}while(isdigit(c)){x=(x<<1)+(x<<3)+c-'0';c=getchar();}return f*x;
}
int n,m;
int ans;
struct node{int to,nxt,from;
}p[N<<1];
int fi[N],cnt;
inline void addline(int x,int y){//printf("%d->%d w=%d\n",x,y,w);p[++cnt]=(node){y,fi[x],x};fi[x]=cnt;
}
int dfn[N],low[N],zhan[N],tim,top,cut[N],r,tot;
void tarjan(int x){dfn[x]=low[x]=++tim;zhan[++top]=x;int nm=0;for(int i=fi[x];~i;i=p[i].nxt){int to=p[i].to;//printf("x=%d to=%d\n",x,to);if(!dfn[to]){++nm;tarjan(to);if(low[to]>=dfn[x]){cut[x]=1;}low[x]=min(low[x],low[to]);}else low[x]=min(low[x],dfn[to]);}if(x==r) cut[x]=nm>1;
}
int num,Cut,tag[N],t;
bool vis[N];
void dfs(int x){if(vis[x]||tag[x]==t) return;num++;//printf("x=%d\n",x);if(cut[x]){tag[x]=t;Cut++;return;}vis[x]=1;for(int i=fi[x];~i;i=p[i].nxt){int to=p[i].to;dfs(to);}
}
int main(){#ifndef ONLINE_JUDGEfreopen("a.in","r",stdin);freopen("a.out","w",stdout);#endifint o=0;while(1){memset(fi,-1,sizeof(fi));cnt=-1;memset(vis,0,sizeof(vis));memset(dfn,0,sizeof(dfn));tim=0;top=0;memset(cut,0,sizeof(cut));memset(tag,0,sizeof(tag));m=read();n=0;if(!m) return 0;for(int i=1;i<=m;i++){int x=read(),y=read();addline(x,y);addline(y,x);n=max(n,max(x,y));}tot=n;for(int i=1;i<=n;i++){r=i;if(!dfn[i]) tarjan(i);}ll nm=0,ans=1;for(int i=1;i<=n;i++){//printf("i=%d cut=%d\n",i,cut[i]);if(!vis[i]&&!cut[i]){num=0;Cut=0;t=i;dfs(i);printf("i=%d num=%d Cut=%d\n",i,num,Cut);if(Cut==0){nm+=2;ans*=num*(num-1)/2;}else if(Cut==1){nm++;ans*=(num-1);}}}printf("Case %d: %lld %lld\n",++o,nm,ans);}
}

边双连通分量

定义：边连通度>1的极大子图

把桥求出后去掉
剩下的连通块就是边双连通分量
加边使图边双连通的方法：把所有桥连回去，形成一个树。设树度数为1的点的数量为$x$
若x=1，不用加边
否则，加边数为$(x+1)/2$

代码

洛谷P2860

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N=5e5+100;
const double eps=1e-6;
inline ll read(){ll x=0,f=1;char c=getchar();while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}while(isdigit(c)){x=(x<<1)+(x<<3)+c-'0';c=getchar();}return f*x;
}
int n,m;
int ans;
struct node{int to,nxt;
}p[N<<1];
int fi[N],cnt;
inline void addline(int x,int y){//printf("%d->%d w=%d\n",x,y,w);p[++cnt]=(node){y,fi[x]};fi[x]=cnt;
}
int low[N],dfn[N],tim,from[N],cut[N],du[N];
int fa[N];
inline int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);
}
inline void merge(int x,int y){x=find(x),y=find(y);fa[x]=y;return;
}
void tarjan(int x){dfn[x]=low[x]=++tim;for(int i=fi[x];~i;i=p[i].nxt){//printf("  x=%d to=%d\n",x,p[i].to);if(i==(from[x]^1)) continue;int to=p[i].to;//printf("  x=%d to=%d\n",x,to);if(!dfn[to]){from[to]=i;tarjan(to);low[x]=min(low[x],low[to]);if(low[to]>dfn[x]){cut[i]=2;cut[i^1]=1;}}low[x]=min(low[x],dfn[to]);}//printf("x=%d dfn=%d low=%d\n",x,dfn[x],low[x]);
}
int main(){#ifndef ONLINE_JUDGEfreopen("a.in","r",stdin);freopen("a.out","w",stdout);#endifmemset(fi,-1,sizeof(fi));cnt=1;n=read();m=read();for(int i=1;i<=m;i++){int x=read(),y=read();addline(x,y);addline(y,x);}for(int i=1;i<=n;i++){if(!dfn[i]) tarjan(i);}for(int i=1;i<=n;i++){fa[i]=i;}for(int x=1;x<=n;x++){for(int i=fi[x];~i;i=p[i].nxt){if(cut[i]) continue;merge(x,p[i].to);}}for(int x=1;x<=n;x++){int xx=find(x);for(int i=fi[x];~i;i=p[i].nxt){if(cut[i]!=2) continue;int yy=find(p[i].to);du[xx]++;du[yy]++;}}for(int i=1;i<=n;i++){if(find(i)==i&&du[i]==1) ans++;}printf("%d\n",ans==1?0:(ans+1)/2);
}
/*
3
intercommunicational
alkylbenzenesulfonate
tetraiodophenolphthalein
0
*/