Problem p01317 (C++) — buggy submission, verdict: Runtime Error

#include <bits/stdc++.h>

#define FOR(i, a, b) for (int i = (a); i < ((int)b); ++i)
#define RFOR(i, a) for (int i = (a); i >= 0; --i)
#define FOE(i, a) for (auto i : a)
#define ALL(c) (c).begin(), (c).end()
#define RALL(c) (c).rbegin(), (c).rend()
#define DUMP(x) cerr << #x << " = " << (x) << endl;
#define SUM(x) std::accumulate(ALL(x), 0LL)
#define MIN(v) *std::min_element(v.begin(), v.end())
#define MAX(v) *std::max_element(v.begin(), v.end())
#define EXIST(v, x) (std::find(v.begin(), v.end(), x) != v.end())
#define BIT(n) (1LL << (n))
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end());

typedef long long LL;
template <typename T> using V = std::vector<T>;
template <typename T> using VV = std::vector<std::vector<T>>;
template <typename T> using VVV = std::vector<std::vector<std::vector<T>>>;

template <class T> inline T ceil(T a, T b) { return (a + b - 1) / b; }
template <class T> inline void print(T x) { std::cout << x << std::endl; }
template <class T> inline void print_vec(const std::vector<T> &v) {
  for (int i = 0; i < v.size(); ++i) {
    if (i != 0) {
      std::cout << " ";
    }
    std::cout << v[i];
  }
  std::cout << "\\n";
}
template <class T> inline bool inside(T y, T x, T H, T W) {
  return 0 <= y and y < H and 0 <= x and x < W;
}
template <class T> inline double euclidean_distance(T y1, T x1, T y2, T x2) {
  return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}
template <class T> inline double manhattan_distance(T y1, T x1, T y2, T x2) {
  return abs(x1 - x2) + abs(y1 - y2);
}

const int INF = 1L << 30;
const double EPS = 1e-9;
const std::string YES = "YES", Yes = "Yes", NO = "NO", No = "No";
const std::vector<int> dy4 = {0, 1, 0, -1},
                       dx4 = {1, 0, -1, 0}; // 4近傍（右, 下, 左, 上）
const std::vector<int> dy8 = {0, -1, 0, 1, 1, -1, -1, 1},
                       dx8 = {1, 0, -1, 0, 1, 1, -1, -1};

using namespace std;

/**
 * 全ノード間の最短距離をもとめる
 * matrix[i][j]には辺e=(i,j)のコスト(辺が存在しない場合はINT_MAX)
 * 負の閉路がある場合は空を返す
 * O(|V|^3)
 */
std::vector<std::vector<LL>>
warshall_floyd(std::vector<std::vector<LL>> matrix) {
  const unsigned long num_node = matrix.size();

  // 自分の距離は0
  for (int i = 0; i < num_node; ++i) {
    matrix.at(i).at(i) = 0;
  }

  for (int m = 0; m < num_node; ++m) {
    for (int s = 0; s < num_node; ++s) {
      for (int e = 0; e < num_node; ++e) {
        // sからeへmを経由して到達可能
        if (matrix.at(s).at(m) != INT_MAX and matrix.at(m).at(e) != INT_MAX) {
          matrix.at(s).at(e) = std::min(
              matrix.at(s).at(e), matrix.at(s).at(m) + matrix.at(m).at(e));
        }
      }
    }
  }

  // 負閉路チェック
  for (int u = 0; u < num_node; ++u) {
    if (matrix.at(u).at(u) < 0) {
      return {};
    }
  }

  return matrix;
}

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  while (true) {
    int N, M;
    cin >> N >> M;
    if (N == 0 and M == 0) {
      break;
    }

    vector<vector<LL>> matrix1(N, vector<LL>(N, INT_MAX)),
        matrix2(N, vector<LL>(N, INT_MAX));
    FOR(i, 0, M) {
      LL X, Y, T;
      char S;
      cin >> X >> Y >> T >> S;
      X--;
      Y--;
      if (S == 'L') {
        matrix1[X][Y] = min(matrix1[X][Y], T);
        matrix1[Y][X] = min(matrix1[Y][X], T);
      } else if (S == 'S') {
        matrix2[X][Y] = min(matrix2[X][Y], T);
        matrix2[Y][X] = min(matrix2[Y][X], T);
      }
    }

    int R;
    cin >> R;
    vector<int> route(R);
    FOR(i, 0, R) {
      cin >> route[i];
      route[i]--;
    }

    matrix1 = warshall_floyd(matrix1);
    matrix2 = warshall_floyd(matrix2);

    vector<vector<LL>> dp(R, vector<LL>(N, INF));
    dp[0][route[0]] = 0;

    FOR(i, 0, R - 1) {
      int now = route[i];
      int next = route[i + 1];

      FOR(j, 0, N) {
        if (dp[i][j] == INF) {
          continue;
        }

        FOR(k, 0, N) {
          // 船がjにあるとき、町nowから町jに陸路、町jから町kに海路、町kから町nextに陸路でいく
          LL dist =
              dp[i][j] + matrix1[now][j] + matrix2[j][k] + matrix1[k][next];
          dp[i + 1][k] = min(dp[i + 1][k], dist);

          if (j == k) {
            LL d = dp[i][j] + matrix1[now][next];
            dp[i + 1][j] = min(dp[i + 1][j], d);
          }
        }
      }
    }

    LL ans = INF;
    FOR(i, 0, N) { ans = min(ans, dp[R - 1][i]); }
    print(ans);
  }

  return 0;
}


Error: 0

Fixed submission:

#include <bits/stdc++.h>

#define FOR(i, a, b) for (int i = (a); i < ((int)b); ++i)
#define RFOR(i, a) for (int i = (a); i >= 0; --i)
#define FOE(i, a) for (auto i : a)
#define ALL(c) (c).begin(), (c).end()
#define RALL(c) (c).rbegin(), (c).rend()
#define DUMP(x) cerr << #x << " = " << (x) << endl;
#define SUM(x) std::accumulate(ALL(x), 0LL)
#define MIN(v) *std::min_element(v.begin(), v.end())
#define MAX(v) *std::max_element(v.begin(), v.end())
#define EXIST(v, x) (std::find(v.begin(), v.end(), x) != v.end())
#define BIT(n) (1LL << (n))
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end());

typedef long long LL;
template <typename T> using V = std::vector<T>;
template <typename T> using VV = std::vector<std::vector<T>>;
template <typename T> using VVV = std::vector<std::vector<std::vector<T>>>;

template <class T> inline T ceil(T a, T b) { return (a + b - 1) / b; }
template <class T> inline void print(T x) { std::cout << x << std::endl; }
template <class T> inline void print_vec(const std::vector<T> &v) {
  for (int i = 0; i < v.size(); ++i) {
    if (i != 0) {
      std::cout << " ";
    }
    std::cout << v[i];
  }
  std::cout << "\\n";
}
template <class T> inline bool inside(T y, T x, T H, T W) {
  return 0 <= y and y < H and 0 <= x and x < W;
}
template <class T> inline double euclidean_distance(T y1, T x1, T y2, T x2) {
  return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}
template <class T> inline double manhattan_distance(T y1, T x1, T y2, T x2) {
  return abs(x1 - x2) + abs(y1 - y2);
}

const int INF = 1L << 30;
const double EPS = 1e-9;
const std::string YES = "YES", Yes = "Yes", NO = "NO", No = "No";
const std::vector<int> dy4 = {0, 1, 0, -1},
                       dx4 = {1, 0, -1, 0}; // 4近傍（右, 下, 左, 上）
const std::vector<int> dy8 = {0, -1, 0, 1, 1, -1, -1, 1},
                       dx8 = {1, 0, -1, 0, 1, 1, -1, -1};

using namespace std;

/**
 * 全ノード間の最短距離をもとめる
 * matrix[i][j]には辺e=(i,j)のコスト(辺が存在しない場合はINT_MAX)
 * 負の閉路がある場合は空を返す
 * O(|V|^3)
 */
std::vector<std::vector<LL>>
warshall_floyd(std::vector<std::vector<LL>> matrix) {
  const unsigned long num_node = matrix.size();

  // 自分の距離は0
  for (int i = 0; i < num_node; ++i) {
    matrix.at(i).at(i) = 0;
  }

  for (int m = 0; m < num_node; ++m) {
    for (int s = 0; s < num_node; ++s) {
      for (int e = 0; e < num_node; ++e) {
        // sからeへmを経由して到達可能
        if (matrix.at(s).at(m) != INT_MAX and matrix.at(m).at(e) != INT_MAX) {
          matrix.at(s).at(e) = std::min(
              matrix.at(s).at(e), matrix.at(s).at(m) + matrix.at(m).at(e));
        }
      }
    }
  }

  // 負閉路チェック
  for (int u = 0; u < num_node; ++u) {
    if (matrix.at(u).at(u) < 0) {
      return {};
    }
  }

  return matrix;
}

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  while (true) {
    int N, M;
    cin >> N >> M;
    if (N == 0 and M == 0) {
      break;
    }

    vector<vector<LL>> matrix1(N, vector<LL>(N, INT_MAX)),
        matrix2(N, vector<LL>(N, INT_MAX));
    FOR(i, 0, M) {
      int X, Y, T;
      char S;
      cin >> X >> Y >> T >> S;
      X--;
      Y--;
      if (S == 'L') {
        matrix1[X][Y] = T;
        matrix1[Y][X] = T;
      } else if (S == 'S') {
        matrix2[X][Y] = T;
        matrix2[Y][X] = T;
      }
    }

    int R;
    cin >> R;
    vector<int> route(R);
    FOR(i, 0, R) {
      cin >> route[i];
      route[i]--;
    }

    matrix1 = warshall_floyd(matrix1);
    matrix2 = warshall_floyd(matrix2);

    vector<vector<LL>> dp(R, vector<LL>(N, INF));
    dp[route[0]][route[0]] = 0;

    FOR(i, 0, R - 1) {
      int now = route[i];
      int next = route[i + 1];

      FOR(j, 0, N) {
        FOR(k, 0, N) {
          // 船がjにあるとき、町nowから町jに陸路、町jから町kに海路、町kから町nextに陸路でいく
          LL dist =
              dp[i][j] + matrix1[now][j] + matrix2[j][k] + matrix1[k][next];
          dp[i + 1][k] = min(dp[i + 1][k], dist);
        }
      }
    }

    LL ans = INF;
    FOR(i, 0, N) { ans = min(ans, dp[R - 1][i]); }
    print(ans);
  }

  return 0;
}