【边双联通分量】hihoCoder#1184 连通性二·边的双连通分量

题面在这里

其实就是边双的裸题

示例程序：

//hihocoder 1184#include<cstdio>#include<cstring>#include<algorithm>#define cl(x,y) memset(x,y,sizeof(x))using namespace std;inline char nc(){    static char buf[100000],*p1=buf,*p2=buf;    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline int red(){    int res=0,f=1;char ch=nc();    while (ch<'0'||'9'<ch) {if (ch=='-') f=-f;ch=nc();}    while ('0'<=ch&&ch<='9') res=(res<<3)+(res<<1)+ch-48,ch=nc();    return res*f;}const int maxn=20005,maxe=200005;int n,e;int tot,son[maxe],lnk[maxn],nxt[maxe];inline void add(int x,int y){    son[++tot]=y;nxt[tot]=lnk[x];lnk[x]=tot;}int times,in[maxn],low[maxn];bool isb[maxe];void tarjan(int x,int fa){    low[x]=in[x]=++times;    for (int j=lnk[x];j;j=nxt[j])     if (!in[son[j]]){        tarjan(son[j],x);        low[x]=min(low[x],low[son[j]]);        if (low[son[j]]>dfn[x]) isb[j]=isb[j^1]=1;     }else if (son[j]!=fa) low[x]=min(low[x],in[son[j]]);}int BCC,bcc[maxn],Min[maxn];void bl(int x){    bcc[x]=BCC;Min[BCC]=min(Min[BCC],x);    for (int j=lnk[x];j;j=nxt[j])     if (!isb[j]&&!bcc[son[j]]) bl(son[j]);}int main(){    n=red(),e=red();tot=1;    for (int i=1,x,y;i<=e;i++) x=red(),y=red(),add(x,y),add(y,x);    for (int i=1;i<=n;i++)     if (!in[i]) tarjan(i,0);    cl(Min,63);    for (int i=1;i<=n;i++)     if (!bcc[i]) BCC++,bl(i);    printf("%d\n",BCC);    for (int i=1;i<=n;i++) printf("%d ",Min[bcc[i]]);    return 0;}