CodeChef Counting important pairs

題意：給定一個N 個點的聯通的無向圖，問有多少種方案，刪除2 條邊之後圖不連通。

題解：隨機化+樹形dp+xor.

具體寫起來有些小細節

/*
  ID:Agreement
  PROG:TYPEAIS
*/
// Invincible
#include <bits/stdc++.h>
#define rep( i , l , r ) for( int i = (l) ; i <= (r) ; ++i )
#define per( i , r , l ) for( int i = (r) ; i >= (l) ; --i )
#define erep( i , u ) for( int i = head[(u)] ; ~i ; i = e[i].nxt )
using namespace std;
using namespace std;
inline int read(){
    int x=0;char ch=getchar();for(;ch<'0'||ch>'9';ch=nc());
    for(;ch<='9'&&ch>='0';x=x*10+ch-48,ch=getchar());return x;
}
#define N 600100
typedef unsigned long long ll;
inline ll rp(){return (1ll*rand())^(1ll*rand())<<15^(1ll*rand())<<30^(1ll*rand())<<45^(1ll*rand())<<60;}
int i,j,m,k,n,x,y,d[N],s,fa[N],cnt,last[N],ct,f[N];ll vl[N],hs[N];
struct edge{int to,next;}e[N<<1];ll ans;
inline void add(int u,int v){
    e[++cnt]=(edge){v,last[u]},last[u]=cnt;
    e[++cnt]=(edge){u,last[v]},last[v]=cnt;
}
inline int find(int x){for(;x!=fa[x];)x=fa[x]=fa[fa[x]];return x;}
void dfs(int x){for(int i=last[x],y;i;i=e[i].next)if((y=e[i].to)!=f[x])f[y]=x,dfs(y),vl[x]^=hs[++ct]=vl[y];}
int main(){
    srand(time(0));
    for(n=read(),i=1;i<=n;++i)fa[i]=i,d[i]=1;
    for(m=read(),i=1;i<=m;++i){
        int p=find(x=read()),q=find(y=read());
        if(p!=q){
            if(add(x,y),d[p]<d[q])swap(p,q);
            if(fa[q]=p,d[p]==d[q])++d[p];
        }else hs[++ct]=rp(),vl[x]^=hs[ct],vl[y]^=hs[ct];
    }
    if(dfs(1),ct!=m)return 1;
    for(sort(hs+1,hs+m+1);s<m&&!hs[s+1];++s);
    for(ans=1ll*s*(m-s),i=1;i<=m;ans+=1ll*(j-i)*(j-i-1)>>1ll,i=j)for(j=i;j<=m&&hs[i]==hs[j];++j);
    printf("%llu\n",ans);
}