Chinese postman problem
(Mylib/Graph/chinese_postman_problem.cpp)

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• Last update: 2021-04-23 23:44:44+09:00
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Operations

• chinese_postman_problem(g)
  • すべての辺を一度以上通る閉路の最小距離を返す。
  • Time complexity $O(n^2 2^n)$

Requirements

Notes

Problems

• AOJ DPL_2_B Chinese Postman Problem

References

• https://ja.wikipedia.org/wiki/%E4%B8%AD%E5%9B%BD%E4%BA%BA%E9%83%B5%E4%BE%BF%E9%85%8D%E9%81%94%E5%95%8F%E9%A1%8C

Depends on

• Basic graph (Mylib/Graph/Template/graph.cpp)

Verified with

• test/aoj/DPL_2_B/main.test.cpp

Code

#pragma once
#include <algorithm>
#include <vector>
#include "Mylib/Graph/Template/graph.cpp"

namespace haar_lib {
  template <typename T>
  T chinese_postman_problem(const graph<T> &g) {
    const int n = g.size();
    T ret       = 0;

    // 頂点間の最短距離を求める。
    std::vector<std::vector<int>> dist(n, std::vector<T>(n, -1));

    for (int i = 0; i < n; ++i) dist[i][i] = 0;

    for (int i = 0; i < n; ++i) {
      for (auto &e : g[i]) {
        if (dist[e.from][e.to] == -1)
          dist[e.from][e.to] = e.cost;
        else
          dist[e.from][e.to] = std::min(dist[e.from][e.to], e.cost);
      }
    }

    for (int k = 0; k < n; ++k) {
      for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
          if (dist[i][k] >= 0 and dist[k][j] >= 0) {
            if (dist[i][j] == -1)
              dist[i][j] = dist[i][k] + dist[k][j];
            else
              dist[i][j] = std::min(dist[i][j], dist[i][k] + dist[k][j]);
          }
        }
      }
    }

    // 奇数次数の頂点を列挙
    std::vector<int> odd;
    for (int i = 0; i < n; ++i) {
      if (g[i].size() % 2) odd.push_back(i);
    }

    const int m = odd.size();

    // 奇数次数の頂点間の最小マッチングを求める。
    std::vector<T> dp(1 << m, -1);
    dp[0] = 0;

    for (int i = 0; i < (1 << m); ++i) {
      for (int j = 0; j < m; ++j) {
        for (int k = 0; k < j; ++k) {
          if ((i & (1 << j)) and (i & (1 << k))) {
            if (dp[i] == -1)
              dp[i] = dp[i ^ (1 << j) ^ (1 << k)] + dist[odd[j]][odd[k]];
            else
              dp[i] = std::min(dp[i], dp[i ^ (1 << j) ^ (1 << k)] + dist[odd[j]][odd[k]]);
          }
        }
      }
    }

    // 返り値を計算
    for (int i = 0; i < n; ++i) {
      for (auto &e : g[i])
        if (e.from <= e.to) ret += e.cost;
    }

    ret += dp[(1 << m) - 1];

    return ret;
  }
}  // namespace haar_lib

#line 2 "Mylib/Graph/chinese_postman_problem.cpp"
#include <algorithm>
#include <vector>
#line 2 "Mylib/Graph/Template/graph.cpp"
#include <iostream>
#line 4 "Mylib/Graph/Template/graph.cpp"

namespace haar_lib {
  template <typename T>
  struct edge {
    int from, to;
    T cost;
    int index = -1;
    edge() {}
    edge(int from, int to, T cost) : from(from), to(to), cost(cost) {}
    edge(int from, int to, T cost, int index) : from(from), to(to), cost(cost), index(index) {}
  };

  template <typename T>
  struct graph {
    using weight_type = T;
    using edge_type   = edge<T>;

    std::vector<std::vector<edge<T>>> data;

    auto& operator[](size_t i) { return data[i]; }
    const auto& operator[](size_t i) const { return data[i]; }

    auto begin() const { return data.begin(); }
    auto end() const { return data.end(); }

    graph() {}
    graph(int N) : data(N) {}

    bool empty() const { return data.empty(); }
    int size() const { return data.size(); }

    void add_edge(int i, int j, T w, int index = -1) {
      data[i].emplace_back(i, j, w, index);
    }

    void add_undirected(int i, int j, T w, int index = -1) {
      add_edge(i, j, w, index);
      add_edge(j, i, w, index);
    }

    template <size_t I, bool DIRECTED = true, bool WEIGHTED = true>
    void read(int M) {
      for (int i = 0; i < M; ++i) {
        int u, v;
        std::cin >> u >> v;
        u -= I;
        v -= I;
        T w = 1;
        if (WEIGHTED) std::cin >> w;
        if (DIRECTED)
          add_edge(u, v, w, i);
        else
          add_undirected(u, v, w, i);
      }
    }
  };

  template <typename T>
  using tree = graph<T>;
}  // namespace haar_lib
#line 5 "Mylib/Graph/chinese_postman_problem.cpp"

namespace haar_lib {
  template <typename T>
  T chinese_postman_problem(const graph<T> &g) {
    const int n = g.size();
    T ret       = 0;

    // 頂点間の最短距離を求める。
    std::vector<std::vector<int>> dist(n, std::vector<T>(n, -1));

    for (int i = 0; i < n; ++i) dist[i][i] = 0;

    for (int i = 0; i < n; ++i) {
      for (auto &e : g[i]) {
        if (dist[e.from][e.to] == -1)
          dist[e.from][e.to] = e.cost;
        else
          dist[e.from][e.to] = std::min(dist[e.from][e.to], e.cost);
      }
    }

    for (int k = 0; k < n; ++k) {
      for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
          if (dist[i][k] >= 0 and dist[k][j] >= 0) {
            if (dist[i][j] == -1)
              dist[i][j] = dist[i][k] + dist[k][j];
            else
              dist[i][j] = std::min(dist[i][j], dist[i][k] + dist[k][j]);
          }
        }
      }
    }

    // 奇数次数の頂点を列挙
    std::vector<int> odd;
    for (int i = 0; i < n; ++i) {
      if (g[i].size() % 2) odd.push_back(i);
    }

    const int m = odd.size();

    // 奇数次数の頂点間の最小マッチングを求める。
    std::vector<T> dp(1 << m, -1);
    dp[0] = 0;

    for (int i = 0; i < (1 << m); ++i) {
      for (int j = 0; j < m; ++j) {
        for (int k = 0; k < j; ++k) {
          if ((i & (1 << j)) and (i & (1 << k))) {
            if (dp[i] == -1)
              dp[i] = dp[i ^ (1 << j) ^ (1 << k)] + dist[odd[j]][odd[k]];
            else
              dp[i] = std::min(dp[i], dp[i ^ (1 << j) ^ (1 << k)] + dist[odd[j]][odd[k]]);
          }
        }
      }
    }

    // 返り値を計算
    for (int i = 0; i < n; ++i) {
      for (auto &e : g[i])
        if (e.from <= e.to) ret += e.cost;
    }

    ret += dp[(1 << m) - 1];

    return ret;
  }
}  // namespace haar_lib

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