无向图求边双连通分量/桥

在无向图中，去掉了i号边，连通块数量增加了，那么i号边就是桥。

#include
using namespace std;
#define ls rt<<1
#define rs (rt<<1)+1
#define PI acos(-1)
#define eps 1e-8
#define ll long long
#define fuck(x) cout<<#x<<"     "< pii;
const int inf=2e9;
const int maxn=1e6+10;
int d[4][2]={1,0,-1,0,0,1,0,-1};
//int lowbit(int x){return x&-x;}
//void add(int x,int v){while(x<=n)bit[x]+=v,x+=lowbit(x);}
//int sum(int x){int ans=0;while(x>=1) ans+=bit[x],x-=lowbit(x);return ans;}
inline ll read() {
    ll s = 0,w = 1;
    char ch = getchar();
    while(!isdigit(ch)) {
        if(ch == '-') w = -1;
        ch = getchar();
    }
    while(isdigit(ch))
        s = s * 10 + ch - '0',ch = getchar();
    return s * w;
}
inline void write(ll x) {
    if(x < 0)
        putchar('-'), x = -x;
    if(x > 9)
        write(x / 10);
    putchar(x % 10 + '0');
}
int gcd(int x,int y){
    return y==0?x:gcd(y,x%y);
}
struct edge{
    int u,v,nxt;
}e[maxn<<1];
int head[maxn],dfn[maxn],low[maxn],cnt,n,m;
vectorbri;

void dfs(int now,int fa){
    bool flag=0;
    dfn[now]=low[now]=++cnt;
    for(int i=head[now];i!=-1;i=e[i].nxt){
        int v=e[i].v;
        if(!dfn[v]){
            dfs(v,now);
            low[now]=min(low[now],low[v]);
            if(low[v]>dfn[now]){
                bri.push_back(i);
            }
        }
        else
            if(v!=fa||flag)
                low[now]=min(low[now],dfn[v]);
            else
                flag=1;
    }
}

int main(){

    return 0;
}

一个边双连通分量里任意两点有至少两条边不重复路径，即没有桥

 

#include
using namespace std;
#define ls rt<<1
#define rs (rt<<1)+1
#define PI acos(-1)
#define eps 1e-8
#define ll long long
#define fuck(x) cout<<#x<<"     "< pii;
const int inf=2e9;
const int maxn=1e6+10;
int d[4][2]={1,0,-1,0,0,1,0,-1};
//int lowbit(int x){return x&-x;}
//void add(int x,int v){while(x<=n)bit[x]+=v,x+=lowbit(x);}
//int sum(int x){int ans=0;while(x>=1) ans+=bit[x],x-=lowbit(x);return ans;}
inline ll read() {
    ll s = 0,w = 1;
    char ch = getchar();
    while(!isdigit(ch)) {
        if(ch == '-') w = -1;
        ch = getchar();
    }
    while(isdigit(ch))
        s = s * 10 + ch - '0',ch = getchar();
    return s * w;
}
inline void write(ll x) {
    if(x < 0)
        putchar('-'), x = -x;
    if(x > 9)
        write(x / 10);
    putchar(x % 10 + '0');
}
int gcd(int x,int y){
    return y==0?x:gcd(y,x%y);
}
struct edge{
    int u,v,nxt;
}e[maxn<<1];
int head[maxn],dfn[maxn],low[maxn],bcc[maxn],bccn,cnt,n,m;
vectorbri;
stacksta;

void dfs(int now,int fa){
    bool flag=0;//flag用于处理重边的情况
    dfn[now]=low[now]=++cnt;
    sta.push(now);
    for(int i=head[now];i!=-1;i=e[i].nxt){
        int v=e[i].v;
        if(!dfn[v]){
            dfs(v,now);
            low[now]=min(low[now],low[v]);
        }
        else
            if(v!=fa||flag)
                low[now]=min(low[now],dfn[v]);
            else
                flag=1;
    }
    if(low[now]==dfn[now]){
        bccn++;
        for(;;){
            int tmp=sta.top();sta.pop();
            bcc[tmp]=bccn;
            if(tmp==now) break;
        }
    }
}

int main(){

    return 0;
}