이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#define FAST ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);cerr.tie(0)
#define mp make_pair
#define xx first
#define yy second
#define pb push_back
#define pf push_front
#define popb pop_back
#define popf pop_front
#define all(x) x.begin(),x.end()
#define ff(i,a,b) for (int i = a; i < b; i++)
#define fff(i,a,b) for (int i = a; i <= b; i++)
#define bff(i,a,b) for (int i = b-1; i >= a; i--)
#define bfff(i,a,b) for (int i = b; i >= a; i--)
using namespace std;
long double typedef ld;
unsigned int typedef ui;
long long int typedef li;
pair<int,int> typedef pii;
pair<li,li> typedef pli;
pair<ld,ld> typedef pld;
vector<vector<int>> typedef graph;
unsigned long long int typedef ull;
//const int mod = 998244353;
const int mod = 1000000007;
//Note to self: Check for overflow
const double eps=1e-9;
vector<pii> w1,w2;
vector<vector<pii>> g(100005); //(gde, boja)
int znak[100005],shift[100005]; //x - c, ili mozda x+c, takodje proverava posecenost ako znak[]=0
ld ans[100005];
bool determined=false; //da li x vec mora biti postavljeno jednacinama?
double x; //ako je determined, sta je ona?
vector<int> poseceni;
void dfs(int p)
{
poseceni.pb(p);
for (auto it : g[p])
{
if (znak[it.xx])
{
if (znak[it.xx]==znak[p]){
//x + shift[p] + x + shift[it.xx] = it.yy
double tx=(double)(it.yy-shift[p]-shift[it.xx])*znak[p]/2;
if (determined){
if (abs(tx-x)>eps) cout<<"NO\n",exit(0);
}
else x=tx,determined=true;
}
else if (shift[p]+shift[it.xx]!=it.yy)
{
cout<<"NO\n",exit(0);
}
}
else
{
znak[it.xx]=-znak[p];
shift[it.xx]=it.yy-shift[p];
dfs(it.xx);
}
}
}
void solve(int root)
{
poseceni.clear();
determined=false;
znak[root]=1,dfs(root);
if (!determined)
{
vector<int> smene;
for (auto i : poseceni) smene.pb(-znak[i]*shift[i]);
sort(all(smene));
int n=(int)smene.size();
x=smene[n/2];
}
for (auto it : poseceni) ans[it]=shift[it]+znak[it]*x;
}
int main()
{
FAST;
int n,m; cin>>n>>m;
while (m--)
{
int u,v,w; cin>>u>>v>>w;
g[u].pb({v,w}),g[v].pb({u,w});
if (w==1) w1.pb({u,v});
else w2.pb({u,v});
}
fff(i,1,n) if (!znak[i]) solve(i);
cout<<"YES\n";
fff(i,1,n) cout<<ans[i]<<" ";
}
//Note to self: Check for overflow
/*
7 12 2
1 2 1
1 3 1
1 4 1
1 5 1
1 6 1
1 7 1
2 3 2
3 4 2
4 5 2
5 6 2
6 7 2
7 2 2
8 10
2 7 1
7 8 1
8 6 1
2 3 2
3 6 2
1 2 2
5 6 2
4 3 1
1 4 1
4 5 1
*/
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