Submission #580083

#TimeUsernameProblemLanguageResultExecution timeMemory
580083jiahngGraph (BOI20_graph)C++14
34 / 100
56 ms8100 KiB
#include <bits/stdc++.h> using namespace std; typedef long long ll; #define int ll typedef pair<int32_t, int32_t> pi; typedef vector <int> vi; typedef vector <pi> vpi; typedef pair<pi, ll> pii; typedef set <ll> si; typedef long double ld; #define f first #define s second #define mp make_pair #define FOR(i,s,e) for(int i=s;i<=int(e);++i) #define DEC(i,s,e) for(int i=s;i>=int(e);--i) #define pb push_back #define all(x) (x).begin(), (x).end() #define lbd(x, y) lower_bound(all(x), y) #define ubd(x, y) upper_bound(all(x), y) #define aFOR(i,x) for (auto i: x) #define mem(x,i) memset(x,i,sizeof x) #define fast ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0) #define maxn 100010 #define INF (ll)1e9+10 #define MOD 1000000007 typedef pair <vi, int> pvi; typedef pair <int,pi> ipi; typedef vector <pii> vpii; int N,M,a,b,c; bool vis[maxn]; vpi adj[maxn]; ld value; int status = 0; // status is 0 if not set, 1 if fixed and -1 if no solution pi val[maxn]; vi v; ld ans[maxn]; void dfs(int x){ if (status == -1) return; vis[x] = 1; v.pb(x); aFOR(i,adj[x]){ if (vis[i.f]){ pi cur = val[x]; cur.f += val[i.f].f; cur.s += val[i.f].s; if (cur.f == 0){ if (cur.s != i.s){ status = -1; return; } }else{ ld newvalue = (ld(i.s) - ld(cur.s)) / ld(cur.f); if (status == 1 && value != newvalue){ status = -1; return; } value = newvalue; status = 1; } }else{ val[i.f] = pi(-val[x].f, i.s - val[x].s); dfs(i.f); } } } int32_t main(){ fast; cin >> N >> M; FOR(i,1,M){ cin >> a >> b >> c; adj[a].pb(pi(b, c)); adj[b].pb(pi(a,c)); } FOR(i,1,N) if (!vis[i]){ val[i] = pi(1,0); status = 0; v.clear(); dfs(i); if (status == -1){ cout << "NO"; return 0; }else if (status == 0){ aFOR(j,v) aFOR(k,adj[j]) assert(val[j].f + val[k.f].f == 0); vi v2; aFOR(j,v){ if (val[j].f == -1) v2.pb(val[j].s); else v2.pb(-val[j].s); } sort(all(v2)); value = v2[v2.size()/2]; } aFOR(j,v) ans[j] = (ld)val[j].f * value + (ld)val[j].s; //aFOR(j,v) aFOR(k,adj[j]) assert(ans[j] + ans[k.f] == k.s); } cout << "YES\n"; FOR(i,1,N) cout << ans[i] << ' '; }
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