結果

問題 No.488 四角関係
ユーザー noynote
提出日時 2017-02-24 22:51:24
言語 C++11
(gcc 13.3.0)
結果
AC  
実行時間 13 ms / 5,000 ms
コード長 2,359 bytes
コンパイル時間 1,672 ms
コンパイル使用メモリ 161,460 KB
実行使用メモリ 6,820 KB
最終ジャッジ日時 2025-01-02 23:30:07
合計ジャッジ時間 2,713 ms
ジャッジサーバーID
(参考情報)
judge4 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 22
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<bits/stdc++.h>
#include<algorithm>
#define range(i,a,b) for(int i = (a); i < (b); i++)
#define rep(i,b) for(int i = 0; i < (b); i++)
#define all(a) (a).begin(), (a).end()
#define show(x) cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;
const int INF = 2000000000;
using namespace std;
template < class BidirectionalIterator >
bool next_combination ( BidirectionalIterator first1 ,
BidirectionalIterator last1 ,
BidirectionalIterator first2 ,
BidirectionalIterator last2 ){
if (( first1 == last1 ) || ( first2 == last2 )) {
return false ;
}
BidirectionalIterator m1 = last1 ;
BidirectionalIterator m2 = last2 ; --m2;
while (--m1 != first1 && !(* m1 < *m2 )){
}
bool result = (m1 == first1 ) && !(* first1 < *m2 );
if (! result ) {
while ( first2 != m2 && !(* m1 < * first2 )) {
++ first2 ;
}
first1 = m1;
std :: iter_swap (first1 , first2 );
++ first1 ;
++ first2 ;
}
if (( first1 != last1 ) && ( first2 != last2 )) {
m1 = last1 ; m2 = first2 ;
while (( m1 != first1 ) && (m2 != last2 )) {
std :: iter_swap (--m1 , m2 );
++ m2;
}
std :: reverse (first1 , m1 );
std :: reverse (first1 , last1 );
std :: reverse (m2 , last2 );
std :: reverse (first2 , last2 );
}
return ! result ;
}
template < class BidirectionalIterator > bool next_combination ( BidirectionalIterator first , BidirectionalIterator middle , BidirectionalIterator
    last )
{
return next_combination (first , middle , middle , last );
}
int main(){
int n, m;
int g[55][55] = {0};
cin >> n >> m;
rep(i,m){
int a, b;
cin >> a >> b;
g[a][b] = g[b][a] = 1;
}
if(n <= 3){
cout << 0 << endl;
return 0;
}
vector<int> v;
int ans = 0;
rep(i,n) v.emplace_back(i);
do{
int sg[4][4];
int cnt = 0;
rep(i,4){
int c = 0;
rep(j,4){
if(g[v[i]][v[j]]) c++;
}
if(c == 2) cnt++;
}
if(cnt == 4) ans++;
}while(next_combination(v.begin(), v.begin() + 4, v.end()));
cout << ans << endl;
}
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