結果
| 問題 |
No.488 四角関係
|
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2019-03-05 01:16:35 |
| 言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
AC
|
| 実行時間 | 214 ms / 5,000 ms |
| コード長 | 4,334 bytes |
| コンパイル時間 | 2,025 ms |
| コンパイル使用メモリ | 180,380 KB |
| 実行使用メモリ | 6,944 KB |
| 最終ジャッジ日時 | 2024-06-23 14:06:39 |
| 合計ジャッジ時間 | 4,146 ms |
|
ジャッジサーバーID (参考情報) |
judge3 / judge2 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 22 |
ソースコード
#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>; using vvi = vector<vi>; using vvvi = vector<vvi>;
using ll = long long int;
using vll = vector<ll>; using vvll = vector<vll>; using vvvll = vector<vvll>;
using vd = vector<double>; using vvd = vector<vd>; using vvvd = vector<vvd>;
using P = pair<int, int>;
using Pll = pair<ll, ll>;
using cdouble = complex<double>;
const double eps = 1e-9;
const double INFD = numeric_limits<double>::infinity();
const double PI = 3.14159265358979323846;
#define Loop(i, n) for(int i = 0; i < (int)(n); i++)
#define Loop1(i, n) for(int i = 1; i <= (int)(n); i++)
#define Loopr(i, n) for(int i = (int)(n) - 1; i >= 0; i--)
#define Loopr1(i, n) for(int i = (int)(n); i >= 1; i--)
#define Foreach(buf, container) for(auto buf : container)
#define Loopdiag(i, j, h, w, sum) for(int i = ((sum) >= (h) ? (h) - 1 : (sum)), j = (sum) - i; i >= 0 && j < (w); i--, j++)
#define Loopdiagr(i, j, h, w, sum) for(int j = ((sum) >= (w) ? (w) - 1 : (sum)), i = (sum) - j; j >= 0 && i < (h); j--, i++)
#define Loopdiagsym(i, j, h, w, gap) for (int i = ((gap) >= 0 ? (gap) : 0), j = i - (gap); i < (h) && j < (w); i++, j++)
#define Loopdiagsymr(i, j, h, w, gap) for (int i = ((gap) > (h) - (w) - 1 ? (h) - 1 : (w) - 1 + (gap)), j = i - (gap); i >= 0 && j >= 0; i--, j--)
#define Loopitr(itr, container) for(auto itr = container.begin(); itr != container.end(); itr++)
#define printv(vector) Loop(ex_i, vector.size()) { cout << vector[ex_i] << " "; } cout << endl;
#define printmx(matrix) Loop(ex_i, matrix.size()) { Loop(ex_j, matrix[ex_i].size()) { cout << matrix[ex_i][ex_j] << " "; } cout << endl; }
#define quickio() ios::sync_with_stdio(false); cin.tie(0);
#define bitmanip(m,val) static_cast<bitset<(int)m>>(val)
#define Comp(type_t) bool operator<(const type_t &another) const
#define fst first
#define snd second
bool nearlyeq(double x, double y) { return abs(x - y) < eps; }
bool inrange(int x, int t) { return x >= 0 && x < t; }
bool inrange(vi xs, int t) { Foreach(x, xs) if (!(x >= 0 && x < t)) return false; return true; }
ll rndf(double x) { return (ll)(x + (x >= 0 ? 0.5 : -0.5)); }
ll floorsqrt(ll x) { ll m = (ll)sqrt((double)x); return m + (m * m <= x ? 0 : -1); }
ll ceilsqrt(ll x) { ll m = (ll)sqrt((double)x); return m + (x <= m * m ? 0 : 1); }
ll rnddiv(ll a, ll b) { return (a / b + (a % b * 2 >= b ? 1 : 0)); }
ll ceildiv(ll a, ll b) { return (a / b + (a % b == 0 ? 0 : 1)); }
ll gcd(ll m, ll n) { if (n == 0) return m; else return gcd(n, m % n); }
ll lcm(ll m, ll n) { return m * n / gcd(m, n); }
/*******************************************************/
int n, m;
vvi mx;
vvi perms;
template<typename val_t>
class Partial_Permutation {
private:
int n;
vector<bool> used;
vector<vector<val_t>> result;
vvi facts; // iPj
void core_func(const vector<val_t> &a, int n, int r, int start) {
if (r == 0) return;
int m = facts[n - 1][r - 1];
int cnt = 0;
Loop(i, Partial_Permutation::n) {
if (!used[i]) {
Loop(j, m) {
result[start + m * cnt + j].push_back(a[i]);
}
used[i] = true;
core_func(a, n - 1, r - 1, start + m * cnt);
used[i] = false;
cnt++;
}
}
}
void make_facts(int n) {
facts = vvi(n + 1, vi(n + 1));
Loop(i, n + 1) {
facts[i][0] = 1;
Loop(j, i) {
facts[i][j + 1] = facts[i][j] * (i - j);
}
}
}
public:
vector<vector<val_t>> get_partial_permutation(const vector<val_t> &a, int r) {
n = int(a.size());
if (n < r) return {};
used = vector<bool>(n, false);
make_facts(n);
result = vector<vector<val_t>>(facts[n][r]);
core_func(a, n, r, 0);
return result;
}
};
int solve(vi a) {
Foreach(ary, perms) {
bool judge = true;
Loop(i, 4) {
if (!mx[a[ary[i]]][a[ary[(i + 1) & 0b11]]]) judge = false;
}
if (mx[a[ary[0]]][a[ary[2]]] || mx[a[ary[1]]][a[ary[3]]]) {
judge = false;
}
if (judge) return 1;
}
return 0;
}
int main() {
cin >> n >> m;
mx = vvi(n, vi(n));
Loop(i, m) {
int s, t; cin >> s >> t;
mx[s][t] = 1;
mx[t][s] = 1;
}
int ans = 0;
Partial_Permutation<int> pp;
perms = pp.get_partial_permutation({ 0,1,2,3 }, 4);
Loop(i0, n) {
Loop(i1, n) {
if (i1 <= i0) continue;
Loop(i2, n) {
if (i2 <= i1) continue;
Loop(i3, n) {
if (i3 <= i2) continue;
ans += solve({ i0,i1,i2,i3 });
}
}
}
}
cout << ans << endl;
}