제출 #548926

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
548926		leaked	철인 이종 경기 (APIO18_duathlon)	C++17	0 / 100	91 ms	26920 KiB

count_triplets

#include <bits/stdc++.h>

#define f first
#define s second
#define m_p make_pair
#define vec vector
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define sz(x) (int)(x).size()
#define pw(x) (1LL<<(x))
#define fast_prep ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
typedef pair<int,int> pii;
typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> pll;
typedef unsigned long long ull;
//#define int long long
template<class T> bool umin(T &a,const T &b){return (a>b?a=b,1:0);}
template<class T> bool umax(T &a,const T &b){return (a<b?a=b,1:0);}
const int N=1e5+1;

struct dsu{
    int p[N];
    int cnt[N];
    dsu(){
        iota(p,p+N,0);
        fill(cnt,cnt+N,1);
    }
    int get(int v){
        return p[v]=(p[v]==v?v:get(p[v]));
    }
    void mg(int a,int b){
        a=get(a),b=get(b);
        if(a==b) return;
        p[a]=b;cnt[b]+=cnt[a];
    }
}xy;
vec<int> gr[N],g[N];
int sub[N];
int in[N],fup[N],tt=1;
int cnt=0;
vec<pii> e;
void dfs1(int v,int p){
    in[v]=fup[v]=tt++;
    ++cnt;
    for(auto &z : g[v]){
        if(z==p) continue;
        if(in[z]){
            umin(fup[v],in[z]);
            continue;
        }
        dfs1(z,v);
        umin(fup[v],fup[z]);
        if(fup[z]<=in[v]) xy.mg(v,z);
        else e.pb({v,z});
//        cout<<"WOT "<<(fup[z]<=in[v])<<' '<<v<<' '<<z<<endl;
    }
}
ll dp[N][2];
ll up[N][2];
ll wow[N];
void dfs2(int v,int p){
    dp[v][0]+=xy.cnt[v];
    ll have=1;
    for(auto &z : gr[v]){
        dfs2(z,v);
        dp[v][0]+=dp[z][0];
        dp[v][1]+=dp[z][1];
//        dp[v][1]+=wow[z];
        dp[v][1]+=2ll*dp[z][0]*have;
        have+=dp[z][0];
    }
//    cout<<"WOW "<<v<<' '<<dp[v][1]<<endl;
}
ll bad=0;
void dfs3(int v){
    bad+=dp[v][1]+up[v][1];
//    cout<<
    ll have=cnt-dp[v][0]+xy.cnt[v];
    bad+=2ll*(xy.cnt[v]-1)*(cnt-dp[v][0]);
    ll par=2ll*(cnt-dp[v][0])*xy.cnt[v];
//    cout<<"WAT "<<v<<' '<<bad<<endl;
    for(auto &z : gr[v]){
        par+=2ll*dp[z][0]*have;
        have+=dp[z][0];
        bad+=2ll*(xy.cnt[v]-1)*(dp[z][0]);
    }
//    cout<<"WOW "<<par<<' '<<up[v][1]<<' '<<have<<endl;
    for(auto &z : gr[v]){
        up[z][1]=par-(2ll*(have-dp[z][0])*(dp[z][0]))+up[v][1];
        dfs3(z);
    }
}
signed main(){
    fast_prep;
    int n,m;
    cin>>n>>m;
    for(int i=0;i<m;i++){
        int v,u;
        cin>>v>>u;
        g[v].pb(u);g[u].pb(v);
    }
//    for(int i=1;i<=n;i++) dp[i][0]=1;
    ll total=0;

    for(int i=1;i<=n;i++){
        if(!in[i]){
            dfs1(i,i);

            for(auto &z : e){
//                cout<<"G "<<xy.get(z.f)<<' '<<xy.get(z.s)<<endl;
                gr[xy.get(z.f)].pb(xy.get(z.s));
            }
            dfs2(xy.get(i),xy.get(i));
            dfs3(xy.get(i));
//            cout<<cnt<<''endl;
            total+=1ll*cnt*(cnt-1)*(cnt-2)-bad;
            bad=0;
            cnt=0;e.clear();
        }
    }
    cout<<total;
    return 0;
}
/*
abbaaa
cbbbbccbbccbbbbbbc
(((((((()))())))))
5 5
4 2
2 5
4 1
2 1
4 5

*/

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