結果
| 問題 |
No.1065 電柱 / Pole (Easy)
|
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2020-05-29 21:53:54 |
| 言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
CE
(最新)
AC
(最初)
|
| 実行時間 | - |
| コード長 | 3,800 bytes |
| コンパイル時間 | 1,889 ms |
| コンパイル使用メモリ | 197,808 KB |
| 最終ジャッジ日時 | 2025-01-10 17:03:46 |
|
ジャッジサーバーID (参考情報) |
judge3 / judge4 |
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コンパイルエラー時のメッセージ・ソースコードは、提出者また管理者しか表示できないようにしております。(リジャッジ後のコンパイルエラーは公開されます)
ただし、clay言語の場合は開発者のデバッグのため、公開されます。
ただし、clay言語の場合は開発者のデバッグのため、公開されます。
コンパイルメッセージ
main.cpp: In instantiation of ‘std::istream& operator>>(std::istream&, std::vector<_Tp>&) [with T = std::pair<long double, long double>; std::istream = std::basic_istream<char>]’:
main.cpp:54:42: required from here
main.cpp:35:94: error: no match for ‘operator>>’ (operand types are ‘std::istream’ {aka ‘std::basic_istream<char>’} and ‘std::pair<long double, long double>’)
35 | template<class T>inline istream& operator>>(istream& is, vector<T>& v) { for (auto& a : v)is >> a; return is; }
| ~~~^~~~
In file included from /usr/include/c++/13/sstream:40,
from /usr/include/c++/13/complex:45,
from /usr/include/c++/13/ccomplex:39,
from /usr/include/x86_64-linux-gnu/c++/13/bits/stdc++.h:127,
from main.cpp:5:
/usr/include/c++/13/istream:325:7: note: candidate: ‘std::basic_istream<_CharT, _Traits>::__istream_type& std::basic_istream<_CharT, _Traits>::operator>>(void*&) [with _CharT = char; _Traits = std::char_traits<char>; __istream_type = std::basic_istream<char>]’
325 | operator>>(void*& __p)
| ^~~~~~~~
/usr/include/c++/13/istream:325:25: note: no known conversion for argument 1 from ‘std::pair<long double, long double>’ to ‘void*&’
325 | operator>>(void*& __p)
| ~~~~~~~^~~
/usr/include/c++/13/istream:224:7: note: candidate: ‘std::basic_istream<_CharT, _Traits>::__istream_type& std::basic_istream<_CharT, _Traits>::operator>>(long double&) [with _CharT = char; _Traits = std::char_traits<char>; __istream_type = std::basic_istream<char>]’
224 | operator>>(long double& __f)
| ^~~~~~~~
/usr/include/c++/13/istream:224:31: note: no known conversion for argument 1 from ‘std::pair<long double, long double>’ to ‘long double&’
224 | operator>>(long double& __f)
| ~~~~~~~~~~
ソースコード
//四則演算 #pragma GCC target("avx")
//並列計算 #pragma GCC optimize("O3")
//条件分岐を減らす #pragma GCC optimize("unroll-loops")
//浮動小数点演算 #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using VI = vector<int>;
using VL = vector<ll>;
using VS = vector<string>;
template<class T> using PQ = priority_queue<T, vector<T>, greater<T>>;
#define FOR(i,a,n) for(int i=(a);i<(n);++i)
#define eFOR(i,a,n) for(int i=(a);i<=(n);++i)
#define rFOR(i,a,n) for(int i=(n)-1;i>=(a);--i)
#define erFOR(i,a,n) for(int i=(n);i>=(a);--i)
#define each(i, a) for(auto &i : a)
#define SORT(i) sort(i.begin(),i.end())
#define rSORT(i) sort(i.rbegin(),i.rend())
#define fSORT(i,a) sort(i.begin(),i.end(),a)
#define all(i) i.begin(),i.end()
#define out(y,x) ((y)<0||h<=(y)||(x)<0||w<=(x))
#define line cout << "-----------------------------\n"
#define ENDL(i,n) ((i) == (n) - 1 ? "\n" : " ")
#define stop system("pause")
constexpr ll INF = 1000000000;
constexpr ll LLINF = 1LL << 60;
constexpr ll mod = 1000000007;
constexpr ll MOD = 998244353;
constexpr ld eps = 1e-10;
constexpr ld pi = 3.1415926535897932;
template<class T>inline bool chmax(T& a, const T& b) { if (a < b) { a = b; return true; }return false; }
template<class T>inline bool chmin(T& a, const T& b) { if (a > b) { a = b; return true; }return false; }
inline void init() { cin.tie(nullptr); cout.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(15); }
template<class T>inline istream& operator>>(istream& is, vector<T>& v) { for (auto& a : v)is >> a; return is; }
template<class T>inline istream& operator>>(istream& is, deque<T>& v) { for (auto& a : v)is >> a; return is; }
template<class T, class U>inline istream& operator>>(istream& is, pair<T, U>& p) { is >> p.first >> p.second; return is; }
template<class T>inline vector<T> vec(size_t a) { return vector<T>(a); }
template<class T>inline vector<T> defvec(T def, size_t a) { return vector<T>(a, def); }
template<class T, class... Ts>inline auto vec(size_t a, Ts... ts) { return vector<decltype(vec<T>(ts...))>(a, vec<T>(ts...)); }
template<class T, class... Ts>inline auto defvec(T def, size_t a, Ts... ts) { return vector<decltype(defvec<T>(def, ts...))>(a, defvec<T>(def, ts...)); }
inline void print() { cout << "\n"; }
template<class T, class... Ts>inline void print(const T& a, const Ts&... ts) { cout << a << " "; print(ts...); }
template<class T>inline void print(const vector<T>& v) { for (int i = 0; i < v.size(); ++i)cout << v[i] << (i == v.size() - 1 ? "\n" : " "); }
template<class T>inline void print(const vector<vector<T>>& v) { for (auto& a : v)print(a); }
inline string reversed(const string& s) { string t = s; reverse(all(t)); return t; }
int main() {
init();
int n, m; cin >> n >> m;
int s, t; cin >> s >> t;
--s, --t;
vector<pair<ld, ld>> pole(n); cin >> pole;
vector<vector<pair<int, ld>>> g(n);
FOR(i, 0, m) {
int p, q; cin >> p >> q;
--p, --q;
ld cost = sqrtl((pole[p].first - pole[q].first) * (pole[p].first - pole[q].first) + (pole[p].second - pole[q].second) * (pole[p].second - pole[q].second));
g[p].emplace_back(q, cost);
g[q].emplace_back(p, cost);
}
vector<ld> dis(n, LLINF);
dis[s] = 0;
PQ<pair<ld, int>> dik;
dik.emplace(0, s);
while (!dik.empty()) {
ld d; int now;
tie(d, now) = dik.top();
dik.pop();
if (dis[now] < d)continue;
each(dest, g[now]) {
int to; ld cost;
tie(to, cost) = dest;
if (!chmin(dis[to], d + cost))continue;
dik.emplace(d + cost, to);
}
}
print(dis[t]);
return 0;
}