牛ゲー(最短路問題の双対)
(shortest-path/dual-of-shortest-path.hpp)

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• Last update: 2026-06-08 17:59:24+09:00
• Include: #include "shortest-path/dual-of-shortest-path.hpp"

Depends on

• グラフテンプレート (graph/graph-template.hpp)

Verified with

• verify/verify-aoj-other/aoj-0304.test.cpp

Code

#pragma once

#include "../graph/graph-template.hpp"

template <typename T>
struct Dual_of_Shortest_Path {
  int N;
  vector<edge<T>> es;

  Dual_of_Shortest_Path(int _n) : N(_n) {}

  // add constraint f(j) <= f(i) + w
  void add_edge(int i, int j, T c) { es.emplace_back(i, j, c); }

  // if s != -1, solve max{f(t) - f(s)} for each t
  // if unsatisfiable, return empty vector
  vector<T> solve(int start = -1) {
    T INF = numeric_limits<T>::max() / 2.1;
    vector<T> d(N, INF);
    if (start == -1) fill(begin(d), end(d), T{0});
    if (start != -1) d[start] = 0;
    for (int loop = 0; loop < N; loop++) {
      int upd = 0;
      for (auto e : es) {
        if (d[e.src] + e.cost < d[e.to]) {
          d[e.to] = d[e.src] + e.cost;
          upd = 1;
        }
      }
      if (!upd) break;
      if (loop == N - 1) return {};
    }
    return d;
  }
};

/**
 * @brief 牛ゲー(最短路問題の双対)
 */

#line 2 "shortest-path/dual-of-shortest-path.hpp"

#line 2 "graph/graph-template.hpp"

template <typename T>
struct edge {
  int src, to;
  T cost;

  edge(int _to, T _cost) : src(-1), to(_to), cost(_cost) {}
  edge(int _src, int _to, T _cost) : src(_src), to(_to), cost(_cost) {}

  edge &operator=(const int &x) {
    to = x;
    return *this;
  }

  operator int() const { return to; }
};
template <typename T>
using Edges = vector<edge<T>>;
template <typename T>
using WeightedGraph = vector<Edges<T>>;
using UnweightedGraph = vector<vector<int>>;

// Input of (Unweighted) Graph
UnweightedGraph graph(int N, int M = -1, bool is_directed = false,
                      bool is_1origin = true) {
  UnweightedGraph g(N);
  if (M == -1) M = N - 1;
  for (int _ = 0; _ < M; _++) {
    int x, y;
    cin >> x >> y;
    if (is_1origin) x--, y--;
    g[x].push_back(y);
    if (!is_directed) g[y].push_back(x);
  }
  return g;
}

// Input of Weighted Graph
template <typename T>
WeightedGraph<T> wgraph(int N, int M = -1, bool is_directed = false,
                        bool is_1origin = true) {
  WeightedGraph<T> g(N);
  if (M == -1) M = N - 1;
  for (int _ = 0; _ < M; _++) {
    int x, y;
    cin >> x >> y;
    T c;
    cin >> c;
    if (is_1origin) x--, y--;
    g[x].emplace_back(x, y, c);
    if (!is_directed) g[y].emplace_back(y, x, c);
  }
  return g;
}

// Input of Edges
template <typename T>
Edges<T> esgraph([[maybe_unused]] int N, int M, int is_weighted = true,
                 bool is_1origin = true) {
  Edges<T> es;
  for (int _ = 0; _ < M; _++) {
    int x, y;
    cin >> x >> y;
    T c;
    if (is_weighted)
      cin >> c;
    else
      c = 1;
    if (is_1origin) x--, y--;
    es.emplace_back(x, y, c);
  }
  return es;
}

// Input of Adjacency Matrix
template <typename T>
vector<vector<T>> adjgraph(int N, int M, T INF, int is_weighted = true,
                           bool is_directed = false, bool is_1origin = true) {
  vector<vector<T>> d(N, vector<T>(N, INF));
  for (int _ = 0; _ < M; _++) {
    int x, y;
    cin >> x >> y;
    T c;
    if (is_weighted)
      cin >> c;
    else
      c = 1;
    if (is_1origin) x--, y--;
    d[x][y] = c;
    if (!is_directed) d[y][x] = c;
  }
  return d;
}

/**
 * @brief グラフテンプレート
 */
#line 4 "shortest-path/dual-of-shortest-path.hpp"

template <typename T>
struct Dual_of_Shortest_Path {
  int N;
  vector<edge<T>> es;

  Dual_of_Shortest_Path(int _n) : N(_n) {}

  // add constraint f(j) <= f(i) + w
  void add_edge(int i, int j, T c) { es.emplace_back(i, j, c); }

  // if s != -1, solve max{f(t) - f(s)} for each t
  // if unsatisfiable, return empty vector
  vector<T> solve(int start = -1) {
    T INF = numeric_limits<T>::max() / 2.1;
    vector<T> d(N, INF);
    if (start == -1) fill(begin(d), end(d), T{0});
    if (start != -1) d[start] = 0;
    for (int loop = 0; loop < N; loop++) {
      int upd = 0;
      for (auto e : es) {
        if (d[e.src] + e.cost < d[e.to]) {
          d[e.to] = d[e.src] + e.cost;
          upd = 1;
        }
      }
      if (!upd) break;
      if (loop == N - 1) return {};
    }
    return d;
  }
};

/**
 * @brief 牛ゲー(最短路問題の双対)
 */

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