제출 #372937

#제출 시각아이디문제언어결과실행 시간메모리
372937ACmachineGraph (BOI20_graph)C++17
17 / 100
2 ms492 KiB
#include <bits/stdc++.h> using namespace std; #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> using namespace __gnu_pbds; template<typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>; template<typename T, typename U> using ordered_map = tree<T, U, less<T>, rb_tree_tag, tree_order_statistics_node_update>; typedef long long ll; typedef long double ld; typedef pair<int,int> pii; typedef pair<ll,ll> pll; typedef vector<int> vi; typedef vector<ll> vll; #define FOR(i,j,k,in) for(int i=(j); i < (k);i+=in) #define FORD(i,j,k,in) for(int i=(j); i >=(k);i-=in) #define REP(i,b) FOR(i,0,b,1) #define REPD(i,b) FORD(i,b,0,1) #define pb push_back #define mp make_pair #define ff first #define ss second #define all(x) begin(x), end(x) #define MANY_TESTS int tcase; cin >> tcase; while(tcase--) const double EPS = 1e-9; const int MOD = 1e9+7; const ll INFF = 1e18; const int INF = 1e9; const ld PI = acos((ld)-1); const vi dy = {1, 0, -1, 0, -1, 1, 1, -1}; const vi dx = {0, 1, 0, -1, -1, 1, -1, 1}; void DBG(){cout << "]" << endl;} template<typename T, typename ...U> void DBG(const T& head, const U... args){ cout << head << "; "; DBG(args...); } #define dbg(...) cout << "Line(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__); #define chk() cout << "Check at line(" << __LINE__ << ") hit." << endl; template<class T, unsigned int U> ostream& operator<<(ostream& out, const array<T, U> &v){out << "["; REP(i, U) out << v[i] << ", "; out << "]"; return out;} template <class T, class U> ostream& operator<<(ostream& out, const pair<T, U> &par) {out << "[" << par.first << ";" << par.second << "]"; return out;} template <class T> ostream& operator<<(ostream& out, const set<T> &cont) { out << "{"; for( const auto &x:cont) out << x << ", "; out << "}"; return out; } template <class T, class U> ostream& operator<<(ostream& out, const map<T, U> &cont) {out << "{"; for( const auto &x:cont) out << x << ", "; out << "}"; return out; } template<class T> ostream& operator<<(ostream& out, const vector<T> &v){ out << "["; REP(i, v.size()) out << v[i] << ", "; out << "]"; return out;} template<class T> istream& operator>>(istream& in, vector<T> &v){ for(auto &x : v) in >> x; return in; } int main(){ ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int n, m; cin >> n >> m; vector<vector<array<int, 2>>> g(n); vector<array<int, 3>> edges(m); REP(i, m){ int a, b, c; cin >> a >> b >> c; a--; b--; g[a].pb({b, 2 * c}); g[b].pb({a, 2 * c}); edges[i] = {a, b, 2 * c}; } vector<int> comp(n, -1); function<void(int, int)> dfs = [&](int v, int id){ comp[v] = id; for(auto x : g[v]) { if(comp[x[0]] == -1) dfs(x[0], id); } }; int curr = 0; REP(i, n){ if(comp[i] == -1){ dfs(i, curr); curr++; } } vector<vector<array<int, 3>>> component_edges(curr); REP(i, m){ component_edges[comp[edges[i][0]]].pb(edges[i]); } vector<vector<int>> component_vertices(curr); REP(i, n) component_vertices[comp[i]].pb(i); vector<int> repr(curr, -1); REP(i, n) repr[comp[i]] = i; vector<int> ans(n, INF); auto go = [&](int s, int val, bool removing){ ans[s] = val; queue<int> q; q.push(s); while(!q.empty()){ int v = q.front(); q.pop(); for(auto x : g[v]){ if(ans[x[0]] == INF){ ans[x[0]] = x[1] - ans[v]; q.push(x[0]); } } } bool correct = true; int co = comp[s]; for(auto e : component_edges[co]){ if(ans[e[0]] + ans[e[1]] != e[2]) correct = false; } int res = 0; for(int x : component_vertices[co]) res += abs(ans[x]); if(removing){ for(int x : component_vertices[co]) ans[x] = INF; } if(!correct) return INF; return res; }; bool possible = true; REP(i, curr){ int mini_val = INF; int mini_res = INF; FOR(j, -8, 9, 1){ int tmp_res = go(repr[i], j, true); if(tmp_res < mini_res){ mini_res = tmp_res; mini_val = j; } } if(mini_res == INF) possible = false; go(repr[i], mini_val, false); } if(!possible){ cout << "NO" << "\n"; } else{ cout << "YES" << "\n"; cout << fixed << setprecision(5) ; REP(i, n){ cout << (double)((double)ans[i] / 2.0) << (i == n - 1 ? "\n" : " "); } } return 0; }
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