#864002 (C++14) No.2286 Join Hands

提出ソース

結果

問題	No.2286 Join Hands
ユーザー	👑 hos.lyric
提出日時	2023-04-28 22:59:57
言語	C++14 (gcc 12.3.0 + boost 1.83.0)
結果	TLE
実行時間	-
コード長	6,194 bytes
コンパイル時間	1,262 ms
コンパイル使用メモリ	118,540 KB
実行使用メモリ	15,808 KB
最終ジャッジ日時	2024-11-17 21:57:24
合計ジャッジ時間	91,274 ms
ジャッジサーバーID （参考情報）	judge2 / judge4

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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	2 ms 10,496 KB
testcase_01	AC	2 ms 10,168 KB
testcase_02	AC	1 ms 10,496 KB
testcase_03	AC	2 ms 10,440 KB
testcase_04	AC	2 ms 10,496 KB
testcase_05	AC	2 ms 10,440 KB
testcase_06	AC	1 ms 10,496 KB
testcase_07	AC	2 ms 10,168 KB
testcase_08	AC	715 ms 10,496 KB
testcase_09	AC	940 ms 10,448 KB
testcase_10	TLE	-
testcase_11	TLE	-
testcase_12	TLE	-
testcase_13	TLE	-
testcase_14	TLE	-
testcase_15	TLE	-
testcase_16	TLE	-
testcase_17	AC	604 ms 10,560 KB
testcase_18	TLE	-
testcase_19	TLE	-
testcase_20	TLE	-
testcase_21	TLE	-
testcase_22	TLE	-
testcase_23	TLE	-
testcase_24	TLE	-
testcase_25	TLE	-
testcase_26	TLE	-
testcase_27	TLE	-
testcase_28	AC	2 ms 10,496 KB
testcase_29	AC	2 ms 10,964 KB
testcase_30	AC	2 ms 10,496 KB
testcase_31	AC	2 ms 10,840 KB
testcase_32	AC	2 ms 10,496 KB
testcase_33	AC	1 ms 10,840 KB
testcase_34	AC	2 ms 10,496 KB
testcase_35	AC	2 ms 10,444 KB
testcase_36	AC	1 ms 10,496 KB
testcase_37	AC	2 ms 10,840 KB
testcase_38	AC	2 ms 10,496 KB
testcase_39	AC	2 ms 10,712 KB
testcase_40	AC	2 ms 10,496 KB
testcase_41	AC	2 ms 10,844 KB
testcase_42	AC	2 ms 10,496 KB
testcase_43	AC	2 ms 10,844 KB
testcase_44	AC	2 ms 10,496 KB
testcase_45	AC	1 ms 5,248 KB
testcase_46	AC	1 ms 5,248 KB
testcase_47	TLE	-
testcase_48	TLE	-
testcase_49	TLE	-
testcase_50	TLE	-
testcase_51	AC	1,944 ms 5,248 KB
testcase_52	AC	1,047 ms 5,248 KB
testcase_53	TLE	-
testcase_54	AC	897 ms 5,248 KB
testcase_55	TLE	-
testcase_56	TLE	-
testcase_57	TLE	-
testcase_58	TLE	-
testcase_59	TLE	-
testcase_60	TLE	-

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ソースコード

raw source code

#include <cassert>
#include <cmath>
#include <cstdint>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <functional>
#include <iostream>
#include <limits>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>

using namespace std;

using Int = long long;

template <class T1, class T2> ostream &operator<<(ostream &os, const pair<T1, T2> &a) { return os << "(" << a.first << ", " << a.second << ")"; };
template <class T> ostream &operator<<(ostream &os, const vector<T> &as) { const int sz = as.size(); os << "["; for (int i = 0; i < sz; ++i) { if (i >= 256) { os << ", ..."; break; } if (i > 0) { os << ", "; } os << as[i]; } return os << "]"; }
template <class T> void pv(T a, T b) { for (T i = a; i != b; ++i) cerr << *i << " "; cerr << endl; }
template <class T> bool chmin(T &t, const T &f) { if (t > f) { t = f; return true; } return false; }
template <class T> bool chmax(T &t, const T &f) { if (t < f) { t = f; return true; } return false; }


// Minimum cost flow by successive shortest paths.
// Assumes that there exists no negative-cost cycle.
// TODO: Check the range of intermediate values.
template <class Flow, class Cost> struct MinCostFlow {
  // Watch out when using types other than int and long long.
  static constexpr Flow FLOW_EPS = 1e-10L;
  static constexpr Flow FLOW_INF = std::numeric_limits<Flow>::max();
  static constexpr Cost COST_EPS = 1e-10L;
  static constexpr Cost COST_INF = std::numeric_limits<Cost>::max();

  int n, m;
  vector<int> ptr, nxt, zu;
  vector<Flow> capa;
  vector<Cost> cost;

  explicit MinCostFlow(int n_) : n(n_), m(0), ptr(n_, -1) {}
  void ae(int u, int v, Flow w, Cost c) {
    assert(0 <= u); assert(u < n);
    assert(0 <= v); assert(v < n);
    assert(0 <= w);
    nxt.push_back(ptr[u]); zu.push_back(v); capa.push_back(w); cost.push_back( c); ptr[u] = m++;
    nxt.push_back(ptr[v]); zu.push_back(u); capa.push_back(0); cost.push_back(-c); ptr[v] = m++;
  }

  vector<Cost> pot, dist;
  vector<bool> vis;
  vector<int> pari;

  // cost slopes[j] per flow when flows[j] <= flow <= flows[j + 1]
  vector<Flow> flows;
  vector<Cost> slopes;

  // Finds a shortest path from s to t in the residual graph.
  // O((n + m) log m) time.
  //   Assumes that the members above are set.
  //   The distance to a vertex might not be determined if it is >= dist[t].
  //   You can pass t = -1 to find a shortest path to each vertex.
  void shortest(int s, int t) {
    using Entry = pair<Cost, int>;
    priority_queue<Entry, vector<Entry>, std::greater<Entry>> que;
    for (int u = 0; u < n; ++u) { dist[u] = COST_INF; vis[u] = false; }
    for (que.emplace(dist[s] = 0, s); !que.empty(); ) {
      const Cost c = que.top().first;
      const int u = que.top().second;
      que.pop();
      if (vis[u]) continue;
      vis[u] = true;
      if (u == t) return;
      for (int i = ptr[u]; ~i; i = nxt[i]) if (capa[i] > FLOW_EPS) {
        const int v = zu[i];
        if (!vis[v]) {
          const Cost cc = c + cost[i] + pot[u] - pot[v];
          if (dist[v] > cc) { que.emplace(dist[v] = cc, v); pari[v] = i; }
        }
      }
    }
  }

  // Finds a minimum cost flow from s to t of amount min{(max flow), limFlow}.
  //   Bellman-Ford takes O(n m) time, or O(m) time if there is no negative-cost
  //   edge, or cannot stop if there exists a negative-cost cycle.
  //   min{(max flow), limFlow} shortest paths if Flow is an integral type.
  pair<Flow, Cost> run(int s, int t, Flow limFlow = FLOW_INF) {
    assert(0 <= s); assert(s < n);
    assert(0 <= t); assert(t < n);
    assert(s != t);
    assert(0 <= limFlow);
    pot.assign(n, 0);
    for (; ; ) {
      bool upd = false;
      for (int i = 0; i < m; ++i) if (capa[i] > FLOW_EPS) {
        const int u = zu[i ^ 1], v = zu[i];
        const Cost cc = pot[u] + cost[i];
        if (pot[v] > cc + COST_EPS) { pot[v] = cc; upd = true; }
      }
      if (!upd) break;
    }
    dist.resize(n);
    vis.resize(n);
    pari.resize(n);
    Flow flow = 0;
    Cost cost = 0;
    flows.clear(); flows.push_back(0);
    slopes.clear();
    for (; flow < limFlow; ) {
      shortest(s, t);
      if (!vis[t]) break;
      for (int u = 0; u < n; ++u) pot[u] += min(dist[u], dist[t]);
      Flow f = limFlow - flow;
      for (int v = t; v != s; ) {
        const int i = pari[v]; if (f > capa[i]) { f = capa[i]; } v = zu[i ^ 1];
      }
      for (int v = t; v != s; ) {
        const int i = pari[v]; capa[i] -= f; capa[i ^ 1] += f; v = zu[i ^ 1];
      }
      flow += f;
      cost += f * (pot[t] - pot[s]);
      flows.push_back(flow);
      slopes.push_back(pot[t] - pot[s]);
    }
    return make_pair(flow, cost);
  }
};

////////////////////////////////////////////////////////////////////////////////


int N, M;
vector<int> A, B;

int main() {
  for (; ~scanf("%d%d", &N, &M); ) {
    A.resize(M);
    B.resize(M);
    for (int i = 0; i < M; ++i) {
      scanf("%d%d", &A[i], &B[i]);
      --A[i];
      --B[i];
    }
    
    MinCostFlow<int, int> mcf(2 + N + N + N + N);
    auto out = [&](int u) -> int { return 2 + u; };
    auto in = [&](int u) -> int { return 2 + N + u; };
    auto toL = [&](int u) -> int { return 2 + N + N + u; };
    auto toR = [&](int u) -> int { return 2 + N + N + N + u; };
    for (int u = 0; u < N; ++u) {
      mcf.ae(0, out(u), 1, 0);
      mcf.ae(in(u), 1, 1, 0);
      mcf.ae(toL(u), in(u), 1, 0);
      mcf.ae(toR(u), in(u), 1, 0);
      // discard
      if (u - 1 >= 0) mcf.ae(out(u), toL(u - 1), 1, 0);
      if (u + 1 <  N) mcf.ae(out(u), toR(u + 1), 1, 0);
    }
    for (int u = 0; u < N - 1; ++u) {
      mcf.ae(toL(u + 1), toL(u), N, 0);
      mcf.ae(toR(u), toR(u + 1), N, 0);
    }
    // use
    for (int i = 0; i < M; ++i) {
      mcf.ae(out(A[i]), in(B[i]), 1, -1);
      mcf.ae(out(B[i]), in(A[i]), 1, -1);
    }
    
    const auto res = mcf.run(0, 1, N);
// cerr<<"res = "<<res<<endl;
    assert(res.first == N);
    const int ans = -res.second;
    printf("%d\n", 2 * ans - N);
  }
  return 0;
}

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結果

テストケース

ソースコード