이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
// IOI 2021
#include <bits/stdc++.h>
using namespace std;
#define int ll
#define endl '\n'
#define ends ' '
#define die(x) return cout << x << endl, 0
#define all(v) v.begin(), v.end()
#define sz(x) (int)(x.size())
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) { cerr << ends << H; debug_out(T...); }
#define debug(...) cerr << "{" << #__VA_ARGS__ << "}:", debug_out(__VA_ARGS__)
typedef long long ll;
typedef pair<int, int> pii;
const int INF = 1e9;
const ll MOD = 1e6 + 7;
////////////////////////////////////////////////////////////////////
const int N = 1e5 + 5;
int X[N], M[N], A[N], E[N], P[N], up = 1, dw = 1;
vector<int> G[N], C[N], Cyc;
void DFSCyc(int v, int p = 0) {
M[v] = 1;
for (int i : G[v]) {
int u = v ^ E[i];
if (u == p) continue;
if (M[u]) dw = u, up = v;
else P[u] = v, DFSCyc(u, v);
}
}
void DFST(int v, int idx) {
int p = v ^ E[idx];
for (int i : G[v]) {
int u = v ^ E[i];
if (i != idx) DFST(u, i);
}
X[idx] = A[v], A[p] -= A[v], A[v] = 0;
}
int32_t main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
mt19937 Rnd(time(0));
int n, m; cin >> n >> m;
if (n < m) die(0);
int s = 0;
for (int i = 0; i < n; i++) cin >> A[i + 1], s += A[i + 1];
if (s % 2 == 1) die(0);
for (int i = 0; i < m; i++) {
int v, u; cin >> v >> u;
G[v].push_back(i), G[u].push_back(i);
E[i] = v ^ u;
}
DFSCyc(1);
Cyc.push_back(dw);
while (up != dw) dw = P[dw], Cyc.push_back(dw);
if ((sz(Cyc) & 1) == 0) die(0);
for (int v : Cyc) M[v] = 2;
for (int v : Cyc) {
for (int i : G[v]) {
int u = v ^ E[i];
if (M[u] != 2) DFST(u, i);
else C[v].push_back(i);
}
}
if (m == n - 1) {
for (int i = 0; i < m; i++) cout << 2 * X[i] << endl;
return 0;
}
int sum = 0;
for (int i = 0; i < sz(Cyc); i++) {
int v = Cyc[i], u = Cyc[(i + 1) % sz(Cyc)];
if ((v ^ C[v][1]) != u) swap(C[v][0], C[v][1]);
if ((i & 1) == 0) sum += A[v];
else sum -= A[v];
}
X[C[Cyc[0]][0]] = sum / 2;
for (int v : Cyc) X[C[v][1]] = A[v] - X[C[v][0]];
for (int i = 0; i < m; i++) cout << 2 * X[i] << endl;
return 0;
}
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