제출 #252481

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
252481		rqi	철인 이종 경기 (APIO18_duathlon)	C++14	49 / 100	235 ms	66020 KiB

count_triplets

#include <bits/stdc++.h>
using namespace std;
 
typedef long long ll;
typedef long double ld;
typedef double db; 
typedef string str; 

typedef pair<int,int> pi;
typedef pair<ll,ll> pl; 
typedef pair<db,db> pd; 

typedef vector<int> vi; 
typedef vector<ll> vl; 
typedef vector<db> vd; 
typedef vector<str> vs; 
typedef vector<pi> vpi;
typedef vector<pl> vpl; 
typedef vector<pd> vpd; 

#define mp make_pair
#define f first
#define s second
#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define rall(x) (x).rbegin(), (x).rend() 
#define rsz resize
#define ins insert 
#define ft front() 
#define bk back()
#define pf push_front 
#define pb push_back
#define eb emplace_back 
#define lb lower_bound 
#define ub upper_bound 

#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)

const int MOD = 1e9+7; // 998244353;
const int MX = 2e5+5; 
const ll INF = 1e18; 
const ld PI = acos((ld)-1);
const int xd[4] = {1,0,-1,0}, yd[4] = {0,1,0,-1}; 
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count()); 

template<class T> bool ckmin(T& a, const T& b) { 
    return b < a ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) { 
    return a < b ? a = b, 1 : 0; } 
template<class T> void remDup(vector<T>& v) { 
    sort(all(v)); v.erase(unique(all(v)),end(v)); }
int pct(int x) { return __builtin_popcount(x); } 
int bits(int x) { return 31-__builtin_clz(x); } // floor(log2(x)) 
int cdiv(int a, int b) { return a/b+!(a<0||a%b == 0); } // division of a by b rounded up, assumes b > 0 
int fstTrue(function<bool(int)> f, int lo, int hi) {
    hi ++; assert(lo <= hi); // assuming f is increasing
    while (lo < hi) { // find first index such that f is true 
        int mid = (lo+hi)/2; 
        f(mid) ? hi = mid : lo = mid+1; 
    } 
    return lo;
}

// INPUT
template<class A> void re(complex<A>& c);
template<class A, class B> void re(pair<A,B>& p);
template<class A> void re(vector<A>& v);
template<class A, size_t SZ> void re(array<A,SZ>& a);

template<class T> void re(T& x) { cin >> x; }
void re(db& d) { str t; re(t); d = stod(t); }
void re(ld& d) { str t; re(t); d = stold(t); }
template<class H, class... T> void re(H& h, T&... t) { re(h); re(t...); }

template<class A> void re(complex<A>& c) { A a,b; re(a,b); c = {a,b}; }
template<class A, class B> void re(pair<A,B>& p) { re(p.f,p.s); }
template<class A> void re(vector<A>& x) { trav(a,x) re(a); }
template<class A, size_t SZ> void re(array<A,SZ>& x) { trav(a,x) re(a); }

// TO_STRING
#define ts to_string
str ts(char c) { return str(1,c); }
str ts(bool b) { return b ? "true" : "false"; }
str ts(const char* s) { return (str)s; }
str ts(str s) { return s; }
template<class A> str ts(complex<A> c) { 
    stringstream ss; ss << c; return ss.str(); }
str ts(vector<bool> v) { 
    str res = "{"; F0R(i,sz(v)) res += char('0'+v[i]);
    res += "}"; return res; }
template<size_t SZ> str ts(bitset<SZ> b) {
    str res = ""; F0R(i,SZ) res += char('0'+b[i]);
    return res; }
template<class A, class B> str ts(pair<A,B> p);
template<class T> str ts(T v) { // containers with begin(), end()
    bool fst = 1; str res = "{";
    for (const auto& x: v) {
        if (!fst) res += ", ";
        fst = 0; res += ts(x);
    }
    res += "}"; return res;
}
template<class A, class B> str ts(pair<A,B> p) {
    return "("+ts(p.f)+", "+ts(p.s)+")"; }

// OUTPUT
template<class A> void pr(A x) { cout << ts(x); }
template<class H, class... T> void pr(const H& h, const T&... t) { 
    pr(h); pr(t...); }
void ps() { pr("\n"); } // print w/ spaces
template<class H, class... T> void ps(const H& h, const T&... t) { 
    pr(h); if (sizeof...(t)) pr(" "); ps(t...); }

// DEBUG
void DBG() { cerr << "]" << endl; }
template<class H, class... T> void DBG(H h, T... t) {
    cerr << ts(h); if (sizeof...(t)) cerr << ", ";
    DBG(t...); }
#ifdef LOCAL // compile with -DLOCAL
#define dbg(...) cerr << "LINE(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__)
#else
#define dbg(...) 0
#endif

// FILE I/O
void setIn(string s) { freopen(s.c_str(),"r",stdin); }
void setOut(string s) { freopen(s.c_str(),"w",stdout); }
void unsyncIO() { ios_base::sync_with_stdio(0); cin.tie(0); }
void setIO(string s = "") {
    unsyncIO();
    // cin.exceptions(cin.failbit); 
    // throws exception when do smth illegal
    // ex. try to read letter into int
    if (sz(s)) { setIn(s+".in"), setOut(s+".out"); } // for USACO
}

/**
 * Description: Biconnected components of edges. Removing any vertex in BCC
    * doesn't disconnect it. To get block-cut tree, create a bipartite graph
    * with the original vertices on the left and a vertex for each BCC on the right. 
    * Draw edge $u\leftrightarrow v$ if $u$ is contained within the BCC for $v$.
    * Self-loops are not included in any BCC while BCCS of size 1 represent
    * bridges.
 * Time: O(N+M)
 * Source: GeeksForGeeks (corrected)
 * Verification: 
    * USACO December 2017, Push a Box -> https://pastebin.com/yUWuzTH8
    * https://cses.fi/problemset/task/1705/
 */

struct BCC {
    vector<vpi> adj; vpi ed; 
    vector<vi> comps; // edges for each bcc
    int N, ti = 0; vi disc, st; 
    void init(int _N) { N = _N; disc.rsz(N), adj.rsz(N); }
    void ae(int x, int y) { 
        adj[x].eb(y,sz(ed)), adj[y].eb(x,sz(ed)), ed.eb(x,y); }
    int dfs(int x, int p = -1) { // return lowest disc
        int low = disc[x] = ++ti;
        trav(e,adj[x]) if (e.s != p) {
            if (!disc[e.f]) {
                st.pb(e.s); // disc[x] < LOW -> bridge
                int LOW = dfs(e.f,e.s); ckmin(low,LOW); 
                if (disc[x] <= LOW) { // get edges in bcc
                    comps.eb(); vi& tmp = comps.bk; // new bcc
                    for (int y = -1; y != e.s; )
                        tmp.pb(y = st.bk), st.pop_back();
                }
            } else if (disc[e.f] < disc[x]) // back-edge
                ckmin(low,disc[e.f]), st.pb(e.s);
        }
        return low;
    }
    void gen() { F0R(i,N) if (!disc[i]) dfs(i);  }
};


const int mx = 100005;
int n;
vi oadj[mx];
bool ovisited[mx];
vi olist;

void ogenList(int node){
    if(ovisited[node]) return;
    //dbg(node);
    ovisited[node] = 1;
    olist.pb(node);
    for(auto u: oadj[node]){
        ogenList(u);
    }
}

int N;
vi adj[mx];

vi cadj[mx]; //compnumber+N+1 vertices
int sub[mx];

void genSub(int node, int prv = -1){
    for(auto u: cadj[node]){
        if(u == prv) continue;
        genSub(u, node);
        sub[node]+=sub[u];
    }
    if(node <= N) sub[node]++;
}

ll hardSolve(){
    ll ans = 0;
    BCC bcc;
    bcc.init(N+1);
    for(int i = 1; i <= N; i++){
        for(auto u: adj[i]){
            if(u > i){
                bcc.ae(i, u);
            }
        }
    }
    bcc.gen();
    for(int i = 1; i <= sz(bcc.comps)-1+N+1; i++){
        cadj[i].clear();
        sub[i] = 0;
    }

    ans = ll(N)*ll(N-1)*ll(N-2);
    vector<vi> comps;

    //dbg(bcc.ed);
    for(int i = 0; i < sz(bcc.comps); i++){
        //dbg(i, bcc.comps[i]);
        comps.pb(vi{});
        for(auto u: bcc.comps[i]){
            comps.bk.pb(bcc.ed[u].f);
            comps.bk.pb(bcc.ed[u].s);
        }

        remDup(comps.bk);
        //dbg(i, comps.bk);
        for(auto u: comps.bk){
            cadj[i+N+1].pb(u);
            cadj[u].pb(i+N+1);
        }
    }

    // dbg(comps);
    // for(int i = 1; i <= sz(bcc.comps)+N; i++){
    //     dbg(i, cadj[i]);
    // }

    genSub(1);

    // for(int i = 1; i <= N+sz(comps); i++){
    //     dbg(i, sub[i]);
    // }
    for(int i = 0; i < sz(comps); i++){
        vi u = comps[i];
        for(auto x: u){
            ll tr = 0;
            if(sub[x] > sub[i+N+1]){
                tr = sub[1]-sub[i+N+1];
            }
            else{
                tr = sub[x];
            }
            //dbg(i+N+1, x, tr);
            ans-=ll(sz(u)-1)*ll(tr)*ll(tr-1);
        }
    }
    return ans;
}

ll solve(){
    return hardSolve();
}

int main() {
    setIO();
    int m;
    cin >> n >> m;
    for(int i = 1; i <= m; i++){
        int u, v;
        cin >> u >> v;
        oadj[u].pb(v);
        oadj[v].pb(u);
    }
    ll ans = 0;
    for(int i = 1; i <= n; i++){
        if(ovisited[i]) continue;
        //dbg(i);
        olist.clear();
        ogenList(i);
        map<int, int> m;
        for(int i = 0; i < sz(olist); i++){
            m[olist[i]] = i+1;
        }
        N = sz(olist);
        for(int i = 1; i <= N; i++){
            adj[i].clear();
        }
        for(auto u: olist){
            for(auto x: oadj[u]){
                adj[m[u]].pb(m[x]);
            }
        }
        ans+=solve();
    }
    ps(ans);
    // you should actually read the stuff at the bottom
}

/* stuff you should look for
    * int overflow, array bounds
    * special cases (n=1?)
    * do smth instead of nothing and stay organized
    * WRITE STUFF DOWN
*/

컴파일 시 표준 에러 (stderr) 메시지

count_triplets.cpp: In function 'void setIn(std::__cxx11::string)':
count_triplets.cpp:130:31: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)', declared with attribute warn_unused_result [-Wunused-result]
 void setIn(string s) { freopen(s.c_str(),"r",stdin); }
                        ~~~~~~~^~~~~~~~~~~~~~~~~~~~~
count_triplets.cpp: In function 'void setOut(std::__cxx11::string)':
count_triplets.cpp:131:32: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)', declared with attribute warn_unused_result [-Wunused-result]
 void setOut(string s) { freopen(s.c_str(),"w",stdout); }
                         ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~

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• Subtask 2: 11 / 11
• Subtask 3: 0 / 8
• Subtask 4: 10 / 10
• Subtask 5: 0 / 13
• Subtask 6: 15 / 15
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