結果

問題 No.1647 Travel in Mitaru city 2
ユーザー hitonanodehitonanode
提出日時 2021-08-13 21:41:23
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 242 ms / 2,500 ms
コード長 9,508 bytes
コンパイル時間 2,049 ms
コンパイル使用メモリ 156,952 KB
最終ジャッジ日時 2025-01-23 18:33:24
ジャッジサーバーID
(参考情報)
judge2 / judge3
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 3
other AC * 48
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

diff #
プレゼンテーションモードにする

#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) (x)
#define dbgif(cond, x) 0
#endif
// Shortest cycle detection of UNDIRECTED SIMPLE graphs based on 01-BFS
// Assumption: only two types of edges are permitted: weight = 0 or W ( > 0)
// Complexity: O(E)
// Verified: <https://codeforces.com/contest/1325/problem/E>
struct ShortestCycle01 {
const int INF = std::numeric_limits<int>::max() / 2;
int V, E;
int INVALID = -1;
std::vector<std::vector<std::pair<int, int>>> to; // (nxt, weight)
ShortestCycle01() = default;
ShortestCycle01(int V) : V(V), E(0), to(V) {}
void add_edge(int s, int t, int len) {
assert(0 <= s and s < V);
assert(0 <= t and t < V);
assert(len >= 0);
to[s].emplace_back(t, len);
to[t].emplace_back(s, len);
E++;
}
std::vector<int> dist;
std::vector<int> prev;
// Find minimum length simple cycle which passes vertex `v`
// - return: (LEN, (a, b))
// - LEN: length of the shortest cycles if exists, INF otherwise.
// - the cycle consists of vertices [v, ..., prev[prev[a]], prev[a], a, b, prev[b], prev[prev[b]], ..., v]
std::pair<int, std::pair<int, int>> Solve(int v) {
assert(0 <= v and v < V);
dist.assign(V, INF);
dist[v] = 0;
prev.assign(V, -1);
std::deque<std::pair<int, int>> bfsq;
std::vector<std::pair<std::pair<int, int>, int>> add_edge;
bfsq.emplace_back(v, -1);
while (!bfsq.empty()) {
int now = bfsq.front().first, prv = bfsq.front().second;
bfsq.pop_front();
for (auto nxt : to[now])
if (nxt.first != prv) {
if (dist[nxt.first] == INF) {
dist[nxt.first] = dist[now] + nxt.second;
prev[nxt.first] = now;
if (nxt.second)
bfsq.emplace_back(nxt.first, now);
else
bfsq.emplace_front(nxt.first, now);
} else
add_edge.emplace_back(std::make_pair(now, nxt.first), nxt.second);
}
}
int minimum_cycle = INF;
int s = -1, t = -1;
for (auto edge : add_edge) {
int a = edge.first.first, b = edge.first.second;
int L = dist[a] + dist[b] + edge.second;
if (L < minimum_cycle) minimum_cycle = L, s = a, t = b;
}
return std::make_pair(minimum_cycle, std::make_pair(s, t));
}
};
// UnionFind Tree (0-indexed), based on size of each disjoint set
struct UnionFind {
std::vector<int> par, cou;
UnionFind(int N = 0) : par(N), cou(N, 1) { iota(par.begin(), par.end(), 0); }
int find(int x) { return (par[x] == x) ? x : (par[x] = find(par[x])); }
bool unite(int x, int y) {
x = find(x), y = find(y);
if (x == y) return false;
if (cou[x] < cou[y]) std::swap(x, y);
par[y] = x, cou[x] += cou[y];
return true;
}
int count(int x) { return cou[find(x)]; }
bool same(int x, int y) { return find(x) == find(y); }
};
int main() {
int H, W, N;
cin >> H >> W >> N;
const int X = 120001;
ShortestCycle01 graph(X * 2);
UnionFind uf(X * 2);
int xi = -1;
map<pint, int> mp;
vector<int> xs, ys;
REP(i, N) {
int x, y;
cin >> x >> y;
graph.add_edge(x, y + X, 1);
xs.push_back(x);
ys.push_back(y);
mp[pint(x, y + X)] = i;
if (!uf.unite(x, y + X)) {
xi = x;
break;
}
}
if (xi < 0) {
puts("-1");
return 0;
}
auto [len, p] = graph.Solve(xi);
auto [a, b] = p;
vector<int> q;
int h = a;
while (h != xi) {
q.push_back(h);
h = graph.prev[h];
}
q.push_back(h);
reverse(ALL(q));
h = b;
while (h != xi) {
q.push_back(h);
h = graph.prev[h];
}
vector<int> rets;
REP(i, q.size()) {
int a = q[i], b = q[(i + 1) % q.size()];
if (a > b) swap(a, b);
rets.push_back(mp.at(pint(a, b)));
}
dbg(xs);
dbg(ys);
dbg(rets);
if (ys[rets[0]] != ys[rets[1]]) rotate(rets.begin(), rets.begin() + 1, rets.end());
dbg(rets);
cout << rets.size() << '\n';
for (auto x : rets) cout << x + 1 << ' ';
cout << '\n';
}
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