結果

問題 No.1833 Subway Planning
ユーザー milkcoffee
提出日時 2021-12-28 14:47:40
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 916 ms / 4,000 ms
コード長 9,226 bytes
コンパイル時間 4,814 ms
コンパイル使用メモリ 254,984 KB
実行使用メモリ 64,384 KB
最終ジャッジ日時 2024-10-02 07:19:57
合計ジャッジ時間 18,159 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 23
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ソースコード

diff #
プレゼンテーションモードにする

#include <bits/stdc++.h>
#if __has_include(<atcoder/all>)
#include <atcoder/all>
using namespace atcoder;
#endif
using namespace std;
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#define ll long long
#define forin(in, n)           \
    for (ll i = 0; i < n; i++) \
    cin >> in[i]
#define forout(out)                         \
    for (ll i = 0; i < (ll)out.size(); i++) \
    cout << out[i] << endl
#define rep(i, n) for (ll i = 0; i < n; ++i)
#define rep_up(i, a, n) for (ll i = a; i < n; ++i)
#define rep_down(i, a, n) for (ll i = a; i >= n; --i)
#define P pair<ll, ll>
#define pb push_back
#define all(v) v.begin(), v.end()
#define fi first
#define se second
#define vvvvll vector< vector <vector <vector<ll> > > >
#define vvvll vector< vector< vector<ll> > >
#define vvll vector< vector<ll> >
#define vll vector<ll>
#define pqll priority_queue<ll>
#define pqllg priority_queue<ll, vector<ll>, greater<ll>>
template<class T> inline void vin(vector<T>& v) { rep(i, v.size()) cin >> v.at(i); }
template <class T>
using V = vector<T>;
constexpr ll INF = (1ll << 60);
constexpr ll mod = 1000000007;
//constexpr ll mod = 998244353;
constexpr double pi = 3.14159265358979323846;
template <typename T>
inline bool chmax(T &a, T b)
{
    if (a < b)
    {
        a = b;
        return 1;
    }
    return 0;
}
template <typename T>
inline bool chmin(T &a, T b)
{
    if (a > b)
    {
        a = b;
        return 1;
    }
    return 0;
}
template <typename T>
void pt(T val)
{
    cout << val << "\n";
}
template <typename T>
void pt_vll(vector<T> &v)
{
    ll vs = v.size();
    rep(i, vs)
    {
        cout << v[i];
        if (i == vs - 1)
            cout << "\n";
        else
            cout << " ";
    }
}
ll mypow(ll a, ll n)
{
    ll ret = 1;
    if (n == 0)
        return 1;
    if (a == 0)
        return 0;
    rep(i, n)
    {
        if (ret > (ll)(9e18 + 10) / a)
            return -1;
        ret *= a;
    }
    return ret;
}
long long modpow(long long a, long long n, long long mod)
{
    long long res = 1;
    while (n > 0)
    {
        if (n & 1)
            res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}
long long modinv(long long a, long long m)
{
    long long b = m, u = 1, v = 0;
    while (b)
    {
        long long t = a / b;
        a -= t * b;
        swap(a, b);
        u -= t * v;
        swap(u, v);
    }
    u %= m;
    if (u < 0)
        u += m;
    return u;
}
struct mint {
    ll x;
    constexpr mint(ll x = 0) noexcept : x((x % mod + mod) % mod) {}
    constexpr mint& operator+=(const mint& a) noexcept {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    constexpr mint& operator-=(const mint& a) noexcept {
        if ((x += mod - a.x) >= mod) x -= mod;
        return *this;
    }
    constexpr mint& operator*=(const mint& a) noexcept { (x *= a.x) %= mod; return *this; }
    constexpr mint& operator/=(const mint& a) noexcept { return *this *= a.inv(); }
    constexpr mint operator-() const noexcept { return mint(-x); }
    constexpr mint operator+(const mint& a) const noexcept { return mint(*this) += a; }
    constexpr mint operator-(const mint& a) const noexcept { return mint(*this) -= a; }
    constexpr mint operator*(const mint& a) const noexcept { return mint(*this) *= a; }
    constexpr mint operator/(const mint& a) const noexcept { return mint(*this) /= a; }
    constexpr bool operator==(const mint& a) const noexcept { return x == a.x; }
    constexpr bool operator!=(const mint& a) const noexcept { return x != a.x; }
    constexpr mint pow(ll n) const {
        if (n == 0) return 1;
        mint res = pow(n >> 1);
        res *= res;
        if (n & 1) res *= *this;
        return res;
    }
    constexpr mint inv() const { return pow(mod - 2); }
    friend istream& operator>>(istream& is, mint& a) noexcept {
        ll v; is >> v;
        a = mint(v);
        return is;
    }
    friend ostream& operator<<(ostream& os, const mint& a) noexcept {
        return os << a.x;
    }
};
struct UnionFind
{
    vector<ll> par, size;
    UnionFind(ll N) : par(N)
    { //最初はすべてが根であるとして初期化
        size.resize(N, 1);
        for (ll i = 0; i < N; i++)
            par[i] = i;
    }
    ll root(ll x)
    { //データxの木の根を再帰で得る
        if (par[x] == x)
            return x;
        return par[x] = root(par[x]);
    }
    void unite(ll x, ll y)
    { //xとyの木を併合
        ll rx = root(x);
        ll ry = root(y);
        if (rx == ry)
            return; //同じ木にあるときはそのまま
        if (size[rx] > size[ry])
        {
            par[ry] = rx;
            size[rx] += size[ry];
        }
        else
        {
            par[rx] = ry;
            size[ry] += size[rx];
        }
        return;
    }
    bool same(ll x, ll y)
    { //2つのデータx,yが属する木が同じならtrue
        ll rx = root(x);
        ll ry = root(y);
        return rx == ry;
    }
    ll treesize(ll x) { return size[root(x)]; }
};
const int MAX = 2010000;
long long fac[MAX], finv[MAX], inv[MAX];
// テーブルを作る前処理
void COMinit()
{
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (ll i = 2; i < MAX; i++)
    {
        fac[i] = fac[i - 1] * i % mod;
        inv[i] = mod - inv[mod % i] * (mod / i) % mod;
        finv[i] = finv[i - 1] * inv[i] % mod;
    }
}
// 二項係数計算
long long COM(ll n, ll k)
{
    if (n < k)
        return 0;
    if (n < 0 || k < 0)
        return 0;
    return fac[n] * (finv[k] * finv[n - k] % mod) % mod;
}
vector<ll> enum_div(ll n)
{ //約数全列挙
    vector<ll> ret;
    for (ll i = 1; i * i <= n; ++i)
    {
        if (n % i == 0)
        {
            ret.push_back(i);
            if (i * i != n)
            {
                ret.push_back(n / i);
            }
        }
    }
    return ret;
}
void make_prime(vector<ll> &ret, ll n)
{ //素因数分解
    ll x = n;
    for (ll i = 2; i * i <= x; i++)
    {
        while (n % i == 0)
        {
            n /= i;
            ret.push_back(i);
        }
    }
    if (n != 1)
    {
        ret.push_back(n);
    }
    return;
}
vector<bool> prime(1000010, true);
vector<ll> pri(1000010);
vector<bool> isprime(int N)
{ //素数判定
    if (N >= 0)
        prime[0] = false;
    if (N >= 1)
        prime[1] = false;
    for (ll i = 2; i * i <= N; i++)
    {
        if (!prime[i])
        {
            continue;
        }
        for (ll j = i * i; j <= N; j += i)
        {
            if (prime[j])
                pri[j] = i;
            prime[j] = false;
        }
    }
    return prime;
}
map<ll,ll> compression(vector<ll> v){
    map<ll,ll> ret;
    ll cnt = 0;
    sort(v.begin(),v.end());
    for(ll i=0; i<v.size(); i++){
        if(!ret.count(v[i])){
            ret[v[i]] = cnt;
            cnt++;
        }
    }
    return ret;
}
ll mydiv(ll x, ll y){ //割り算切り捨て(0除算注意)
    if(y<0){
        x=-x;
        y=-y;
    }
    if(x<0){
        return (x-y+1)/y;
    }
    else return x/y;
}
//素因数分解: void make_prime(vector<ll> &ret, ll n)
//素数判定　: vector<bool> isprime(int N)
//約数全列挙: vector<ll> enum_div(ll n)
struct Edge {
    long long to;
    long long cost;
};
using Graph = vector<vector<Edge> >;
multiset<ll> X,Y;
ll tmp=INF;
vll dist(200001,-1);
vll dis(200001,-1);
void dfs(Graph &G, ll v){
    dist[v]=0;
    for(auto e:G[v]){
        if(dist[e.to]==0) continue;
        X.insert(e.cost);
        dfs(G,e.to);
    }
}
void dfs2(Graph &G, ll v){
    dis[v]=0;
    ll x,y,z;
    auto itr = X.end();
    itr --; x = *itr;
    itr = Y.begin(); y = *itr;
    itr = Y.end(); itr --; z = *itr;
    chmin(tmp,max(x,z-y));
    for(auto e:G[v]){
        if(dis[e.to]==0) continue;
        X.erase(X.find(e.cost));
        Y.insert(e.cost);
        dfs2(G,e.to);
        X.insert(e.cost);
        Y.erase(Y.find(e.cost));
    }
}
void solve(){
    ll n, m, k, cnt = 0, sum = 0, ans = 0;
    cin>>n;
    ll ma=0;
    if(n<2||n>200000){
        pt(-1);
        return;
    }
    Graph G(n);
    UnionFind uf(n);
    vll A(n-1);
    vll B(n-1);
    vll C(n-1);
    rep(i,n-1){
        ll a,b,c;
        cin>>a>>b>>c;
        if(a<1||a>n||b<1||b>n){
            pt(-1);
            return;
        }
        if(c<1||c>1000000000){
            pt(-1);
            return;
        }
        a--;b--;
        if(uf.same(a,b)){
            pt(-1);
            return;
        }
        uf.unite(a,b);
        A[i]=a;
        B[i]=b;
        C[i]=c;
        if(chmax(ma,c)){
            cnt=i;
        }
    }
    rep(i,n-1){
        if(cnt==i) continue;
        G[A[i]].pb({B[i],C[i]});
        G[B[i]].pb({A[i],C[i]});
    }
    Y.insert(C[cnt]);
    X.insert(0);
    dfs(G,A[cnt]);
    dfs2(G,A[cnt]);
    ans=tmp;
    X.clear();
    Y.clear();
    Y.insert(C[cnt]);
    X.insert(0);
    tmp=INF;
    dfs(G,B[cnt]);
    dfs2(G,B[cnt]);
    chmax(ans,tmp);
    pt(ans);
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    //cout << fixed << setprecision(16);
    //ll T;
    //cin>>T;
    //rep(ca,T)
    solve();
}
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