이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
#define X first
#define Y second
//const ll mod = 1000000007;
const ll mod = 998244353;
void solve()
{
ll n, m;
cin >> n >> m;
if (m>n)
{
cout << 0 << "\n";
return;
}
vector<ll> a(n);
ll W = 0;
for (ll i=0;i<n;i++)
{
cin >> a[i];
W += a[i];
}
if (W%2!=0)
{
cout << 0 << "\n";
return;
}
vector<ll> b(m), d(n);
vector<pair<ll,ll>> e(m);
vector<vector<ll>> g(n);
for (ll i=0;i<m;i++)
{
ll u, v;
cin >> u >> v;
u--, v--;
g[u].push_back(i);
g[v].push_back(i);
e[i] = {u,v};
d[u]++, d[v]++;
}
vector<ll> was(m), ans(m);
set<ll> kek;
for (ll i=0;i<n;i++) if (d[i]==1) kek.insert(i);
while (kek.size()>0)
{
ll u = *kek.begin();
ll w = a[u];
for (ll p: g[u])
{
if (was[p]==1) continue;
was[p] = 1;
d[e[p].X]--;
if (d[e[p].X]==1) kek.insert(e[p].X);
if (d[e[p].X]==0) kek.erase(e[p].X);
d[e[p].Y]--;
if (d[e[p].Y]==1) kek.insert(e[p].Y);
if (d[e[p].Y]==0) kek.erase(e[p].Y);
a[e[p].X] -= w;
a[e[p].Y] -= w;
ans[p] = w;
}
}
ll D = 0;
for (ll i=0;i<n;i++) if (d[i]==2) D++;
if (D==0)
{
for (ll i=0;i<n;i++)
{
if (a[i]!=0)
{
cout << 0 << "\n";
return;
}
}
for (ll i=0;i<m;i++) cout << ans[i]*2 << "\n";
return;
}
if (D%2==0)
{
cout << 0 << "\n";
return;
}
for (ll st=0;st<n;st++)
{
if (d[st]!=2) continue;
vector<ll> used(n);
ll u = st;
vector<ll> A, B;
ll S = 0;
for (ll k=0;k<D;k++)
{
A.push_back(a[u]);
S += a[u];
for (ll p: g[u])
{
if (was[p]==1) continue;
was[p] = 1;
B.push_back(p);
u = u^e[p].X^e[p].Y;
break;
}
}
S /= 2;
for (ll i=2;i<D;i+=2) S -= A[i];
ll L = 0, R = D-1;
ll Q = D;
for (ll i=0;Q>0;i=(i+2)%D,Q--)
{
ans[B[i]] = S;
L = (L+2)%D;
S += A[L];
R = (R+2)%D;
S -= A[R];
}
for (ll i=0;i<m;i++) cout << ans[i] << "\n";
return;
}
}
int main()
{
ios_base::sync_with_stdio(false);
ll T;
T = 1;
//cin >> T;
while (T--) solve();
}
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