結果
問題 | No.1480 Many Complete Graphs |
ユーザー |
|
提出日時 | 2021-04-16 22:55:45 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 158 ms / 2,000 ms |
コード長 | 2,237 bytes |
コンパイル時間 | 1,740 ms |
コンパイル使用メモリ | 110,572 KB |
最終ジャッジ日時 | 2025-01-20 20:30:58 |
ジャッジサーバーID (参考情報) |
judge5 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 57 |
コンパイルメッセージ
main.cpp: In function ‘int main()’: main.cpp:74:18: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result] 74 | scanf("%d", &x); | ~~~~~^~~~~~~~~~
ソースコード
#include <iostream>#include <algorithm>#include <map>#include <set>#include <queue>#include <stack>#include <numeric>#include <bitset>#include <cmath>static const int MOD = 1000000007;using ll = long long;using u32 = unsigned;using u64 = unsigned long long;using namespace std;template<class T> constexpr T INF = ::numeric_limits<T>::max() / 32 * 15 + 208;template <typename T>struct edge {int from, to; T cost;edge(int to, T cost) : from(-1), to(to), cost(cost) {}edge(int from, int to, T cost) : from(from), to(to), cost(cost) {}};template <typename T>vector<T> dijkstra(int s,vector<vector<edge<T>>> &G){auto n = G.size();vector<T> d(n, INF<T>);priority_queue<pair<T, int>,vector<pair<T, int>>,greater<>> Q;d[s] = 0;Q.emplace(0, s);while(!Q.empty()){T cost; int i;tie(cost, i) = Q.top(); Q.pop();if(d[i] < cost) continue;for (auto &&e : G[i]) {auto cost2 = cost + e.cost;if(d[e.to] <= cost2) continue;d[e.to] = cost2;Q.emplace(d[e.to], e.to);}}return d;}template <class T>ostream& operator<<(ostream& os, vector<T> v) {os << "{";for (int i = 0; i < v.size(); ++i) {if(i) os << ", ";os << v[i];}return os << "}";}template <class L, class R>ostream& operator<<(ostream& os, pair<L, R> p) {return os << "{" << p.first << ", " << p.second << "}";}int main() {int n, m;cin >> n >> m;vector<vector<edge<ll>>> G(n+2*m);int id = n;for (int i = 0; i < m; ++i) {int k, c;cin >> k >> c;for (int j = 0; j < k; ++j) {int x;scanf("%d", &x);if(x&1) {G[x-1].emplace_back(id, x/2+1);G[id].emplace_back(x-1, x/2+c);G[id+1].emplace_back(x-1, x/2+c+1);} else {G[x-1].emplace_back(id+1, x/2);G[id].emplace_back(x-1, x/2+c);G[id+1].emplace_back(x-1, x/2+c);}}id += 2;}ll ans = dijkstra(0, G)[n-1];if(ans == INF<ll>) puts("-1");else cout << ans << "\n";return 0;}