結果
問題 | No.3056 Disconnected Coloring |
ユーザー |
👑 ![]() |
提出日時 | 2025-03-14 21:36:48 |
言語 | C++23 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 572 ms / 2,000 ms |
コード長 | 4,372 bytes |
コンパイル時間 | 6,081 ms |
コンパイル使用メモリ | 284,368 KB |
実行使用メモリ | 78,076 KB |
最終ジャッジ日時 | 2025-03-14 21:37:07 |
合計ジャッジ時間 | 16,132 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 2 |
other | AC * 34 |
ソースコード
#define ATCODER #define _USE_MATH_DEFINES #include<stdio.h> #include <bit> #include <cstdint> #include<iostream> #include<fstream> #include<algorithm> #include<vector> #include<string> #include <cassert> #include <numeric> #include <unordered_map> #include <unordered_set> #include <queue> #include <math.h> #include <climits> #include <set> #include <map> #include <list> #include <random> #include <iterator> #include <bitset> #include <chrono> #include <type_traits> using namespace std; using ll = long long; using ld = long double; using pll = pair<ll, ll>; using pdd = pair<ld, ld>; //template<claLR T> using pq = priority_queue<T, vector<T>, greater<T>>; #define FOR(i, a, b) for(ll i=(a); i<(b);i++) #define REP(i, n) for(ll i=0; i<(n);i++) #define ROF(i, a, b) for(ll i=(b-1); i>=(a);i--) #define PER(i, n) for(ll i=n-1; i>=0;i--) #define REPREP(i,j,a,b) for(ll i=0;i<a;i++)for(ll j=0;j<b;j++) #define VV(type) vector< vector<type> > #define VV2(type,n,m,val) vector< vector<type> > val;val.resize(n);for(ll i;i<n;i++)val[i].resize(m) #define vec(type) vector<type> #define VEC(type,n,val) vector<type> val;val.resize(n) #define VL vector<ll> #define VVL vector< vector<ll> > #define VP vector< pair<ll,ll> > #define SZ size() #define all(i) begin(i),end(i) #define SORT(i) sort(all(i)) #define BITI(i) (1<<i) #define BITSET(x,i) x | (ll(1)<<i) #define BITCUT(x,i) x & ~(ll(1)<<i) #define EXISTBIT(x,i) (((x>>i) & 1) != 0) #define ALLBIT(n) (ll(1)<<n-1) #define CHMAX(n,v) n=n<v?v:n #define CHMIN(n,v) n=n>v?v:n #define MP(a,b) make_pair(a,b) #define DET2(x1,y1,x2,y2) (x1)*(y2)-(x2)*(y1) #define DET3(x1,y1,z1,x2,y2,z2,x3,y3,z3) (x1)*(y2)*(z3)+(x2)*(y3)*(z1)+(x3)*(y1)*(z2)-(z1)*(y2)*(x3)-(z2)*(y3)*(x1)-(z3)*(y1)*(x2) #define INC(a) for(auto& v:a)v++; #define DEC(a) for(auto& v:a)v--; #define SQU(x) (x)*(x) #define L0 ll(0) #ifdef ATCODER #include <atcoder/all> using namespace atcoder; using mint = modint1000000007; using mint2 = modint998244353; #endif template<typename T = ll> vector<T> read(size_t n) { vector<T> ts(n); for (size_t i = 0; i < n; i++) cin >> ts[i]; return ts; } template<typename TV, const ll N> void read_tuple_impl(TV&) {} template<typename TV, const ll N, typename Head, typename... Tail> void read_tuple_impl(TV& ts) { get<N>(ts).emplace_back(*(istream_iterator<Head>(cin))); read_tuple_impl<TV, N + 1, Tail...>(ts); } template<typename... Ts> decltype(auto) read_tuple(size_t n) { tuple<vector<Ts>...> ts; for (size_t i = 0; i < n; i++) read_tuple_impl<decltype(ts), 0, Ts...>(ts); return ts; } using val = ll; using func = ll; val op(val a, val b) { return a + b; } val e() { return 0; } val mp(func f, val a) { return a + f; } func comp(func f, func g) { return f + g; } func id() { return 0; } ll g; bool f(val a) { return a < g; } ll di[4] = { 1,0,-1,0 }; ll dj[4] = { 0,1,0,-1 }; ll si[4] = { 0,3,3,0 }; ll sj[4] = { 0,0,3,3 }; //ll di[4] = { -1,-1,1,1 }; //ll dj[4] = { -1,1,-1,1 }; //ll di8[8] = { 0,-1,-1,-1,0,1,1,1 }; //ll dj8[8] = { -1,-1,0,1,1,1,0,-1 }; void solve() { ll n, m; cin >> n >> m; auto [u, v] = read_tuple<ll, ll>(m); DEC(u); DEC(v); if (m % 2) { cout << "-1\n"; return; } // そんなわけないだろと思いつつ mf_graph<ll> mf(n); vector<VP> g(n); REP(i, m) { mf.add_edge(u[i], v[i], 1); mf.add_edge(v[i], u[i], 1); g[u[i]].emplace_back(v[i], i); g[v[i]].emplace_back(u[i], i); if ((u[i] == 0 && v[i] == n - 1) || (u[i] == n - 1 && v[i] == 0)) { cout << -1 << "\n"; return; } } string ans(m, 'R'); ll cnt = 0; ll f = mf.flow(0, n - 1); if (f * 2 > m) { cout << "-1\n"; return; } auto res = mf.min_cut(0); dsu d(n); REP(i, n) { for (auto& [v, e] : g[i]) { if (res[i] != res[v]) { cnt += ans[e] == 'R'; ans[e] = 'B'; d.merge(i, v); } } } if (cnt * 2 > m) { cout << "-1\n"; return; } REP(i, m) { if (cnt * 2 == m) break; if (ans[i] == 'B') continue; if (d.same(0, u[i]) && d.same(n - 1, v[i]) || d.same(0, v[i]) && d.same(n - 1, u[i])) continue; d.merge(u[i], v[i]); ans[i] = 'B'; cnt++; } if (cnt * 2 != m) { cout << "-1\n"; return; } cout << ans << "\n"; return; } int main() { ll t = 1; //cin >> t; while (t--) { solve(); } return 0; }