#787826 (C++23) No.2039 Copy and Avoid

提出ソース
結果

問題	No.2039 Copy and Avoid
ユーザー	hitonanode
提出日時	2022-08-12 21:51:42
言語	C++23 (gcc 12.3.0 + boost 1.83.0)
結果	AC
実行時間	48 ms / 2,000 ms
コード長	21,364 bytes
コンパイル時間	2,861 ms
コンパイル使用メモリ	214,232 KB
実行使用メモリ	12,624 KB
最終ジャッジ日時	2024-09-23 02:14:12
合計ジャッジ時間	4,339 ms
ジャッジサーバーID （参考情報）	judge3 / judge2
このコードへのチャレンジ
（要ログイン）
テストケース

テストケース表示
入力	結果	実行時間実行使用メモリ
testcase_00	AC	44 ms 11,472 KB
testcase_01	AC	11 ms 7,820 KB
testcase_02	AC	11 ms 7,812 KB
testcase_03	AC	11 ms 7,888 KB
testcase_04	AC	10 ms 7,812 KB
testcase_05	AC	12 ms 7,824 KB
testcase_06	AC	11 ms 7,832 KB
testcase_07	AC	11 ms 7,820 KB
testcase_08	AC	11 ms 7,808 KB
testcase_09	AC	11 ms 7,788 KB
testcase_10	AC	10 ms 7,912 KB
testcase_11	AC	10 ms 7,928 KB
testcase_12	AC	11 ms 7,908 KB
testcase_13	AC	44 ms 11,412 KB
testcase_14	AC	44 ms 11,732 KB
testcase_15	AC	48 ms 12,624 KB
testcase_16	AC	46 ms 12,240 KB
testcase_17	AC	11 ms 7,828 KB
testcase_18	AC	12 ms 7,880 KB
testcase_19	AC	11 ms 7,960 KB
testcase_20	AC	12 ms 7,908 KB
testcase_21	AC	10 ms 7,828 KB
testcase_22	AC	10 ms 7,852 KB
testcase_23	AC	10 ms 7,892 KB
testcase_24	AC	11 ms 7,816 KB
testcase_25	AC	11 ms 7,816 KB
testcase_26	AC	10 ms 7,784 KB
testcase_27	AC	10 ms 7,880 KB
testcase_28	AC	10 ms 7,816 KB
testcase_29	AC	10 ms 7,984 KB
testcase_30	AC	10 ms 7,876 KB
権限があれば一括ダウンロードができます
ソースコード

raw source code
#include <algorithm>
#include <array>
#include <bitset>
#include <cassert>
#include <chrono>
#include <cmath>
#include <complex>
#include <deque>
#include <forward_list>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iostream>
#include <limits>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <tuple>
#include <type_traits>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
using namespace std;
using lint = long long;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
struct fast_ios { fast_ios(){ cin.tie(nullptr), ios::sync_with_stdio(false), cout << fixed << setprecision(20); }; } fast_ios_;
#define ALL(x) (x).begin(), (x).end()
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
template <typename T, typename V>
void ndarray(vector<T>& vec, const V& val, int len) { vec.assign(len, val); }
template <typename T, typename V, typename... Args> void ndarray(vector<T>& vec, const V& val, int len, Args... args) { vec.resize(len), for_each(begin(vec), end(vec), [&](T& v) { ndarray(v, val, args...); }); }
template <typename T> bool chmax(T &m, const T q) { return m < q ? (m = q, true) : false; }
template <typename T> bool chmin(T &m, const T q) { return m > q ? (m = q, true) : false; }
int floor_lg(long long x) { return x <= 0 ? -1 : 63 - __builtin_clzll(x); }
template <typename T1, typename T2> pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first + r.first, l.second + r.second); }
template <typename T1, typename T2> pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first - r.first, l.second - r.second); }
template <typename T> vector<T> sort_unique(vector<T> vec) { sort(vec.begin(), vec.end()), vec.erase(unique(vec.begin(), vec.end()), vec.end()); return vec; }
template <typename T> int arglb(const std::vector<T> &v, const T &x) { return std::distance(v.begin(), std::lower_bound(v.begin(), v.end(), x)); }
template <typename T> int argub(const std::vector<T> &v, const T &x) { return std::distance(v.begin(), std::upper_bound(v.begin(), v.end(), x)); }
template <typename T> istream &operator>>(istream &is, vector<T> &vec) { for (auto &v : vec) is >> v; return is; }
template <typename T> ostream &operator<<(ostream &os, const vector<T> &vec) { os << '['; for (auto v : vec) os << v << ','; os << ']'; return os; }
template <typename T, size_t sz> ostream &operator<<(ostream &os, const array<T, sz> &arr) { os << '['; for (auto v : arr) os << v << ','; os << ']'; return os; }
#if __cplusplus >= 201703L
template <typename... T> istream &operator>>(istream &is, tuple<T...> &tpl) { std::apply([&is](auto &&... args) { ((is >> args), ...);}, tpl); return is; }
template <typename... T> ostream &operator<<(ostream &os, const tuple<T...> &tpl) { os << '('; std::apply([&os](auto &&... args) { ((os << args << ','), ...);}, tpl); return os << ')'; }
#endif
template <typename T> ostream &operator<<(ostream &os, const deque<T> &vec) { os << "deq["; for (auto v : vec) os << v << ','; os << ']'; return os; }
template <typename T> ostream &operator<<(ostream &os, const set<T> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <typename T, typename TH> ostream &operator<<(ostream &os, const unordered_set<T, TH> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <typename T> ostream &operator<<(ostream &os, const multiset<T> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa) { os << '(' << pa.first << ',' << pa.second << ')'; return os; }
template <typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp) { os << '{'; for (auto v : mp) os << v.first << "=>" << v.second << ','; os << '}'; return os; }
template <typename TK, typename TV, typename TH> ostream &operator<<(ostream &os, const unordered_map<TK, TV, TH> &mp) { os << '{'; for (auto v : mp) os << v.first << "=>" << v.second << ','; os << '}'; return os; }
#ifdef HITONANODE_LOCAL
const string COLOR_RESET = "\033[0m", BRIGHT_GREEN = "\033[1;32m", BRIGHT_RED = "\033[1;31m", BRIGHT_CYAN = "\033[1;36m", NORMAL_CROSSED = "\033[0;9;37m", RED_BACKGROUND = "\033[1;41m", NORMAL_FAINT = "\033[0;2m";
#define dbg(x) cerr << BRIGHT_CYAN << #x << COLOR_RESET << " = " << (x) << NORMAL_FAINT << " (L" << __LINE__ << ") " << __FILE__ << COLOR_RESET << endl
#define dbgif(cond, x) ((cond) ? cerr << BRIGHT_CYAN << #x << COLOR_RESET << " = " << (x) << NORMAL_FAINT << " (L" << __LINE__ << ") " << __FILE__ << COLOR_RESET << endl : cerr)
#else
#define dbg(x) 0
#define dbgif(cond, x) 0
#endif

// Linear sieve algorithm for fast prime factorization
// Complexity: O(N) time, O(N) space:
// - MAXN = 10^7:  ~44 MB,  80~100 ms (Codeforces / AtCoder GCC, C++17)
// - MAXN = 10^8: ~435 MB, 810~980 ms (Codeforces / AtCoder GCC, C++17)
// Reference:
// [1] D. Gries, J. Misra, "A Linear Sieve Algorithm for Finding Prime Numbers,"
//     Communications of the ACM, 21(12), 999-1003, 1978.
// - https://cp-algorithms.com/algebra/prime-sieve-linear.html
// - https://37zigen.com/linear-sieve/
struct Sieve {
    std::vector<int> min_factor;
    std::vector<int> primes;
    Sieve(int MAXN) : min_factor(MAXN + 1) {
        for (int d = 2; d <= MAXN; d++) {
            if (!min_factor[d]) {
                min_factor[d] = d;
                primes.emplace_back(d);
            }
            for (const auto &p : primes) {
                if (p > min_factor[d] or d * p > MAXN) break;
                min_factor[d * p] = p;
            }
        }
    }
    // Prime factorization for 1 <= x <= MAXN^2
    // Complexity: O(log x)           (x <= MAXN)
    //             O(MAXN / log MAXN) (MAXN < x <= MAXN^2)
    template <class T> std::map<T, int> factorize(T x) const {
        std::map<T, int> ret;
        assert(x > 0 and
               x <= ((long long)min_factor.size() - 1) * ((long long)min_factor.size() - 1));
        for (const auto &p : primes) {
            if (x < T(min_factor.size())) break;
            while (!(x % p)) x /= p, ret[p]++;
        }
        if (x >= T(min_factor.size())) ret[x]++, x = 1;
        while (x > 1) ret[min_factor[x]]++, x /= min_factor[x];
        return ret;
    }
    // Enumerate divisors of 1 <= x <= MAXN^2
    // Be careful of highly composite numbers https://oeis.org/A002182/list
    // https://gist.github.com/dario2994/fb4713f252ca86c1254d#file-list-txt (n, (# of div. of n)):
    // 45360->100, 735134400(<1e9)->1344, 963761198400(<1e12)->6720
    template <class T> std::vector<T> divisors(T x) const {
        std::vector<T> ret{1};
        for (const auto p : factorize(x)) {
            int n = ret.size();
            for (int i = 0; i < n; i++) {
                for (T a = 1, d = 1; d <= p.second; d++) {
                    a *= p.first;
                    ret.push_back(ret[i] * a);
                }
            }
        }
        return ret; // NOT sorted
    }
    // Euler phi functions of divisors of given x
    // Verified: ABC212 G https://atcoder.jp/contests/abc212/tasks/abc212_g
    // Complexity: O(sqrt(x) + d(x))
    template <class T> std::map<T, T> euler_of_divisors(T x) const {
        assert(x >= 1);
        std::map<T, T> ret;
        ret[1] = 1;
        std::vector<T> divs{1};
        for (auto p : factorize(x)) {
            int n = ret.size();
            for (int i = 0; i < n; i++) {
                ret[divs[i] * p.first] = ret[divs[i]] * (p.first - 1);
                divs.push_back(divs[i] * p.first);
                for (T a = divs[i] * p.first, d = 1; d < p.second; a *= p.first, d++) {
                    ret[a * p.first] = ret[a] * p.first;
                    divs.push_back(a * p.first);
                }
            }
        }
        return ret;
    }
    // Moebius function Table, (-1)^{# of different prime factors} for square-free x
    // return: [0=>0, 1=>1, 2=>-1, 3=>-1, 4=>0, 5=>-1, 6=>1, 7=>-1, 8=>0, ...] https://oeis.org/A008683
    std::vector<int> GenerateMoebiusFunctionTable() const {
        std::vector<int> ret(min_factor.size());
        for (unsigned i = 1; i < min_factor.size(); i++) {
            if (i == 1) {
                ret[i] = 1;
            } else if ((i / min_factor[i]) % min_factor[i] == 0) {
                ret[i] = 0;
            } else {
                ret[i] = -ret[i / min_factor[i]];
            }
        }
        return ret;
    }
    // Calculate [0^K, 1^K, ..., nmax^K] in O(nmax)
    // Note: **0^0 == 1**
    template <class MODINT> std::vector<MODINT> enumerate_kth_pows(long long K, int nmax) const {
        assert(nmax < int(min_factor.size()));
        assert(K >= 0);
        if (K == 0) return std::vector<MODINT>(nmax + 1, 1);
        std::vector<MODINT> ret(nmax + 1);
        ret[0] = 0, ret[1] = 1;
        for (int n = 2; n <= nmax; n++) {
            if (min_factor[n] == n) {
                ret[n] = MODINT(n).pow(K);
            } else {
                ret[n] = ret[n / min_factor[n]] * ret[min_factor[n]];
            }
        }
        return ret;
    }
};
Sieve sieve((1 << 20));


template <typename T, T INF = std::numeric_limits<T>::max() / 2, int INVALID = -1>
struct shortest_path {
    int V, E;
    bool single_positive_weight;
    T wmin, wmax;

    std::vector<std::pair<int, T>> tos;
    std::vector<int> head;
    std::vector<std::tuple<int, int, T>> edges;

    void build_() {
        if (int(tos.size()) == E and int(head.size()) == V + 1) return;
        tos.resize(E);
        head.assign(V + 1, 0);
        for (const auto &e : edges) ++head[std::get<0>(e) + 1];
        for (int i = 0; i < V; ++i) head[i + 1] += head[i];
        auto cur = head;
        for (const auto &e : edges) {
            tos[cur[std::get<0>(e)]++] = std::make_pair(std::get<1>(e), std::get<2>(e));
        }
    }

    shortest_path(int V = 0) : V(V), E(0), single_positive_weight(true), wmin(0), wmax(0) {}
    void add_edge(int s, int t, T w) {
        assert(0 <= s and s < V);
        assert(0 <= t and t < V);
        edges.emplace_back(s, t, w);
        ++E;
        if (w > 0 and wmax > 0 and wmax != w) single_positive_weight = false;
        wmin = std::min(wmin, w);
        wmax = std::max(wmax, w);
    }

    void add_bi_edge(int u, int v, T w) {
        add_edge(u, v, w);
        add_edge(v, u, w);
    }

    std::vector<T> dist;
    std::vector<int> prev;

    // Dijkstra algorithm
    // - Requirement: wmin >= 0
    // - Complexity: O(E log E)
    using Pque = std::priority_queue<std::pair<T, int>, std::vector<std::pair<T, int>>,
                                     std::greater<std::pair<T, int>>>;
    template <class Heap = Pque> void dijkstra(int s, int t = INVALID) {
        assert(0 <= s and s < V);
        build_();
        dist.assign(V, INF);
        prev.assign(V, INVALID);
        dist[s] = 0;
        Heap pq;
        pq.emplace(0, s);
        while (!pq.empty()) {
            T d;
            int v;
            std::tie(d, v) = pq.top();
            pq.pop();
            if (t == v) return;
            if (dist[v] < d) continue;
            for (int e = head[v]; e < head[v + 1]; ++e) {
                const auto &nx = tos[e];
                T dnx = d + nx.second;
                if (dist[nx.first] > dnx) {
                    dist[nx.first] = dnx, prev[nx.first] = v;
                    pq.emplace(dnx, nx.first);
                }
            }
        }
    }

    // Dijkstra algorithm
    // - Requirement: wmin >= 0
    // - Complexity: O(V^2 + E)
    void dijkstra_vquad(int s, int t = INVALID) {
        assert(0 <= s and s < V);
        build_();
        dist.assign(V, INF);
        prev.assign(V, INVALID);
        dist[s] = 0;
        std::vector<char> fixed(V, false);
        while (true) {
            int r = INVALID;
            T dr = INF;
            for (int i = 0; i < V; i++) {
                if (!fixed[i] and dist[i] < dr) r = i, dr = dist[i];
            }
            if (r == INVALID or r == t) break;
            fixed[r] = true;
            int nxt;
            T dx;
            for (int e = head[r]; e < head[r + 1]; ++e) {
                std::tie(nxt, dx) = tos[e];
                if (dist[nxt] > dist[r] + dx) dist[nxt] = dist[r] + dx, prev[nxt] = r;
            }
        }
    }

    // Bellman-Ford algorithm
    // - Requirement: no negative loop
    // - Complexity: O(VE)
    bool bellman_ford(int s, int nb_loop) {
        assert(0 <= s and s < V);
        build_();
        dist.assign(V, INF), prev.assign(V, INVALID);
        dist[s] = 0;
        for (int l = 0; l < nb_loop; l++) {
            bool upd = false;
            for (int v = 0; v < V; v++) {
                if (dist[v] == INF) continue;
                for (int e = head[v]; e < head[v + 1]; ++e) {
                    const auto &nx = tos[e];
                    T dnx = dist[v] + nx.second;
                    if (dist[nx.first] > dnx) dist[nx.first] = dnx, prev[nx.first] = v, upd = true;
                }
            }
            if (!upd) return true;
        }
        return false;
    }

    // Bellman-ford algorithm using deque
    // - Requirement: no negative loop
    // - Complexity: O(VE)
    void spfa(int s) {
        assert(0 <= s and s < V);
        build_();
        dist.assign(V, INF);
        prev.assign(V, INVALID);
        dist[s] = 0;
        std::deque<int> q;
        std::vector<char> in_queue(V);
        q.push_back(s), in_queue[s] = 1;
        while (!q.empty()) {
            int now = q.front();
            q.pop_front(), in_queue[now] = 0;
            for (int e = head[now]; e < head[now + 1]; ++e) {
                const auto &nx = tos[e];
                T dnx = dist[now] + nx.second;
                int nxt = nx.first;
                if (dist[nxt] > dnx) {
                    dist[nxt] = dnx;
                    if (!in_queue[nxt]) {
                        if (q.size() and dnx < dist[q.front()]) { // Small label first optimization
                            q.push_front(nxt);
                        } else {
                            q.push_back(nxt);
                        }
                        prev[nxt] = now, in_queue[nxt] = 1;
                    }
                }
            }
        }
    }

    // 01-BFS
    // - Requirement: all weights must be 0 or w (positive constant).
    // - Complexity: O(V + E)
    void zero_one_bfs(int s, int t = INVALID) {
        assert(0 <= s and s < V);
        build_();
        dist.assign(V, INF), prev.assign(V, INVALID);
        dist[s] = 0;
        std::vector<int> q(V * 4);
        int ql = V * 2, qr = V * 2;
        q[qr++] = s;
        while (ql < qr) {
            int v = q[ql++];
            if (v == t) return;
            for (int e = head[v]; e < head[v + 1]; ++e) {
                const auto &nx = tos[e];
                T dnx = dist[v] + nx.second;
                if (dist[nx.first] > dnx) {
                    dist[nx.first] = dnx, prev[nx.first] = v;
                    if (nx.second) {
                        q[qr++] = nx.first;
                    } else {
                        q[--ql] = nx.first;
                    }
                }
            }
        }
    }

    // Dial's algorithm
    // - Requirement: wmin >= 0
    // - Complexity: O(wmax * V + E)
    void dial(int s, int t = INVALID) {
        assert(0 <= s and s < V);
        build_();
        dist.assign(V, INF), prev.assign(V, INVALID);
        dist[s] = 0;
        std::vector<std::vector<std::pair<int, T>>> q(wmax + 1);
        q[0].emplace_back(s, dist[s]);
        int ninq = 1;

        int cur = 0;
        T dcur = 0;
        for (; ninq; ++cur, ++dcur) {
            if (cur == wmax + 1) cur = 0;
            while (!q[cur].empty()) {
                int v = q[cur].back().first;
                T dnow = q[cur].back().second;
                q[cur].pop_back(), --ninq;
                if (v == t) return;
                if (dist[v] < dnow) continue;

                for (int e = head[v]; e < head[v + 1]; ++e) {
                    const auto &nx = tos[e];
                    T dnx = dist[v] + nx.second;
                    if (dist[nx.first] > dnx) {
                        dist[nx.first] = dnx, prev[nx.first] = v;
                        int nxtcur = cur + int(nx.second);
                        if (nxtcur >= int(q.size())) nxtcur -= q.size();
                        q[nxtcur].emplace_back(nx.first, dnx), ++ninq;
                    }
                }
            }
        }
    }

    // Solver for DAG
    // - Requirement: graph is DAG
    // - Complexity: O(V + E)
    bool dag_solver(int s) {
        assert(0 <= s and s < V);
        build_();
        dist.assign(V, INF), prev.assign(V, INVALID);
        dist[s] = 0;
        std::vector<int> indeg(V, 0);
        std::vector<int> q(V * 2);
        int ql = 0, qr = 0;
        q[qr++] = s;
        while (ql < qr) {
            int now = q[ql++];
            for (int e = head[now]; e < head[now + 1]; ++e) {
                const auto &nx = tos[e];
                ++indeg[nx.first];
                if (indeg[nx.first] == 1) q[qr++] = nx.first;
            }
        }
        ql = qr = 0;
        q[qr++] = s;
        while (ql < qr) {
            int now = q[ql++];
            for (int e = head[now]; e < head[now + 1]; ++e) {
                const auto &nx = tos[e];
                --indeg[nx.first];
                if (dist[nx.first] > dist[now] + nx.second)
                    dist[nx.first] = dist[now] + nx.second, prev[nx.first] = now;
                if (indeg[nx.first] == 0) q[qr++] = nx.first;
            }
        }
        return *max_element(indeg.begin(), indeg.end()) == 0;
    }

    // Retrieve a sequence of vertex ids that represents shortest path [s, ..., goal]
    // If not reachable to goal, return {}
    std::vector<int> retrieve_path(int goal) const {
        assert(int(prev.size()) == V);
        assert(0 <= goal and goal < V);
        if (dist[goal] == INF) return {};
        std::vector<int> ret{goal};
        while (prev[goal] != INVALID) {
            goal = prev[goal];
            ret.push_back(goal);
        }
        std::reverse(ret.begin(), ret.end());
        return ret;
    }

    void solve(int s, int t = INVALID) {
        if (wmin >= 0) {
            if (single_positive_weight) {
                zero_one_bfs(s, t);
            } else if (wmax <= 10) {
                dial(s, t);
            } else {
                if ((long long)V * V < (E << 4)) {
                    dijkstra_vquad(s, t);
                } else {
                    dijkstra(s, t);
                }
            }
        } else {
            bellman_ford(s, V);
        }
    }

    // Warshall-Floyd algorithm
    // - Requirement: no negative loop
    // - Complexity: O(E + V^3)
    std::vector<std::vector<T>> floyd_warshall() {
        build_();
        std::vector<std::vector<T>> dist2d(V, std::vector<T>(V, INF));
        for (int i = 0; i < V; i++) {
            dist2d[i][i] = 0;
            for (const auto &e : edges) {
                int s = std::get<0>(e), t = std::get<1>(e);
                dist2d[s][t] = std::min(dist2d[s][t], std::get<2>(e));
            }
        }
        for (int k = 0; k < V; k++) {
            for (int i = 0; i < V; i++) {
                if (dist2d[i][k] == INF) continue;
                for (int j = 0; j < V; j++) {
                    if (dist2d[k][j] == INF) continue;
                    dist2d[i][j] = std::min(dist2d[i][j], dist2d[i][k] + dist2d[k][j]);
                }
            }
        }
        return dist2d;
    }

    void to_dot(std::string filename = "shortest_path") const {
        std::ofstream ss(filename + ".DOT");
        ss << "digraph{\n";
        build_();
        for (int i = 0; i < V; i++) {
            for (int e = head[i]; e < head[i + 1]; ++e) {
                ss << i << "->" << tos[e].first << "[label=" << tos[e].second << "];\n";
            }
        }
        ss << "}\n";
        ss.close();
        return;
    }
};


int main() {
    int N, M, A, B;
    cin >> N >> M >> A >> B;

    auto ds = sieve.divisors(N);
    sort(ds.begin(), ds.end());
    dbg(ds);

    vector<int> C(M);
    cin >> C;

    shortest_path<lint> graph(ds.size());

    REP(s, ds.size()) {
        int v0 = ds.at(s);
        int hi = N;
        for (auto c : C) {
            if (c % v0 == 0) {
                chmin(hi, c - 1);
            }
        }
        if (hi < v0) continue;

        FOR(t, s + 1, ds.size()) {
            int v1 = ds.at(t);
            if (v1 > hi or v1 % v0) continue;
            graph.add_edge(s, t, (v1 / v0 - 1) * lint(A) + B);
        }
    }
    graph.solve(0);

    auto ret = graph.dist.back();
    if (ret > 1LL << 60) {
        puts("-1");
    } else {
        cout << ret - B << endl;
    }
}
yukicoder

結果

テストケース

ソースコード
