結果
問題 | No.630 門松グラフ |
ユーザー |
|
提出日時 | 2018-01-03 12:03:25 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 9 ms / 1,500 ms |
コード長 | 5,377 bytes |
コンパイル時間 | 3,211 ms |
コンパイル使用メモリ | 216,184 KB |
最終ジャッジ日時 | 2025-01-05 06:56:25 |
ジャッジサーバーID (参考情報) |
judge5 / judge1 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 4 |
other | AC * 32 |
コンパイルメッセージ
main.cpp: In member function ‘{anonymous}::MaiScanner& {anonymous}::MaiScanner::operator>>(std::string&)’: main.cpp:55:9: warning: no return statement in function returning non-void [-Wreturn-type] 55 | } | ^
ソースコード
#pragma GCC optimize ("O3")#pragma GCC target ("avx")#include "bits/stdc++.h" // define macro "/D__MAI"using namespace std;typedef long long int ll;#define debugv(v) {printf("L%d %s > ",__LINE__,#v);for(auto e:v){cout<<e<<" ";}cout<<endl;}#define debuga(m,w) {printf("L%d %s > ",__LINE__,#m);for(int x=0;x<(w);x++){cout<<(m)[x]<<" ";}cout<<endl;}#define debugaa(m,h,w) {printf("L%d %s >\n",__LINE__,#m);for(int y=0;y<(h);y++){for(int x=0;x<(w);x++){cout<<(m)[y][x]<<" ";}cout<<endl;}}#define ALL(v) (v).begin(),(v).end()#define repeat(cnt,l) for(auto cnt=0ll;(cnt)<(l);++(cnt))#define rrepeat(cnt,l) for(auto cnt=(l)-1;0<=(cnt);--(cnt))#define iterate(cnt,b,e) for(auto cnt=(b);(cnt)!=(e);++(cnt))#define MD 1000000007ll#define PI 3.1415926535897932384626433832795template<typename T1, typename T2> ostream& operator <<(ostream &o, const pair<T1, T2> p) { o << "(" << p.first << ":" << p.second << ")"; return o;}template<typename T> T& maxset(T& to, const T& val) { return to = max(to, val); }template<typename T> T& minset(T& to, const T& val) { return to = min(to, val); }void bye(string s, int code = 0) { cout << s << endl; exit(0); }mt19937_64 randdev(8901016);inline ll rand_range(ll l, ll h) {return uniform_int_distribution<ll>(l, h)(randdev);}#if defined(_WIN32) || defined(_WIN64)#define getchar_unlocked _getchar_nolock#define putchar_unlocked _putchar_nolock#elif __GNUC__#else#define getchar_unlocked getchar#define putchar_unlocked putchar#endifnamespace {#define isvisiblechar(c) (0x21<=(c)&&(c)<=0x7E)class MaiScanner {public:template<typename T> void input_integer(T& var) {var = 0; T sign = 1;int cc = getchar_unlocked();for (; cc<'0' || '9'<cc; cc = getchar_unlocked())if (cc == '-') sign = -1;for (; '0' <= cc&&cc <= '9'; cc = getchar_unlocked())var = (var << 3) + (var << 1) + cc - '0';var = var*sign;}inline int c() { return getchar_unlocked(); }inline MaiScanner& operator>>(int& var) { input_integer<int>(var); return *this; }inline MaiScanner& operator>>(long long& var) { input_integer<long long>(var); return *this; }inline MaiScanner& operator>>(string& var) {int cc = getchar_unlocked();for (; !isvisiblechar(cc); cc = getchar_unlocked());for (; isvisiblechar(cc); cc = getchar_unlocked())var.push_back(cc);}template<typename IT> void in(IT begin, IT end) { for (auto it = begin; it != end; ++it) *this >> *it; }};class MaiPrinter {public:template<typename T>void output_integer(T var) {if (var == 0) { putchar_unlocked('0'); return; }if (var < 0)putchar_unlocked('-'),var = -var;char stack[32]; int stack_p = 0;while (var)stack[stack_p++] = '0' + (var % 10),var /= 10;while (stack_p)putchar_unlocked(stack[--stack_p]);}inline MaiPrinter& operator<<(char c) { putchar_unlocked(c); return *this; }inline MaiPrinter& operator<<(int var) { output_integer<int>(var); return *this; }inline MaiPrinter& operator<<(long long var) { output_integer<long long>(var); return *this; }inline MaiPrinter& operator<<(const char* str_p) { while (*str_p) putchar_unlocked(*(str_p++)); return *this; }inline MaiPrinter& operator<<(const string& str) {const char* p = str.c_str();const char* l = p + str.size();while (p < l) putchar_unlocked(*p++);return *this;}template<typename IT> void join(IT begin, IT end, char sep='\n') { for (auto it = begin; it != end; ++it) *this << *it << sep; }};}MaiScanner scanner;MaiPrinter printer;// ----------------------------------------------------// 答案ここから// ---------------------------------------------------void connect(int u, int v) {printer << (u + 1) << ' ' << (v + 1) << '\n';}int main() {ll n, m;scanner >> n >> m;// 二部グラフの頂点数na, n-na// 完全二部グラフを考えたとき,辺の数が最も多くなるようなna, nbを求める.(na < nb)ll na = n/2;ll mbest = na*(n-na);// 入力から与えられたmが範囲内かどうか調べる.if (m < n - 1 || mbest < m) {printer << "NO" << '\n';return 0;}printer << "YES \n";// 頂点に書き込まれる値 = 頂点番号とする.for (int i = 0; i < n; ++i) { printer << (i + 1) << ' '; }printer << '\n';// 連結にするための最小限の辺を引く.connect(0, na); --m;for (int i = 1; i < na; ++i) {connect(i, na + i - 1);connect(i, na + i);m -= 2;}for (int i = na*2; i < n; ++i) {connect(na - 1, i);m -= 1;}// 残りの辺を加える.int ai = 0;int bi = 1;while (0 < m) {if (ai != bi && ai != bi + 1) {connect(ai, na + bi);--m;}// インクリメントif (++bi >= n - na) {++ai;bi = 0;}}return 0;}