QOJ.ac
QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#596596 | #7755. Game on a Forest | rxzfn639 | WA | 1ms | 5304kb | C++23 | 1.9kb | 2024-09-28 16:05:40 | 2024-09-28 16:05:42 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
const i64 P = 998244353;
struct DSU {
vector<int> fa, p, e, f;
DSU(int n) {
fa.resize(n + 1);
iota(fa.begin(), fa.end(), 0);
p.resize(n + 1, 1);
e.resize(n + 1);
f.resize(n + 1);
}
int get(int x) {
if (x != fa[x]) {
fa[x] = get(fa[x]);
}
return fa[x];
}
bool merge(int x, int y) { // 设x是y的祖先
if (x == y) f[get(x)] = 1;
x = get(x), y = get(y);
e[x]++;
if (x == y) return false;
if (x < y) swap(x, y); // 将编号小的合并到大的上
fa[y] = x;
f[x] |= f[y], p[x] += p[y], e[x] += e[y];
return true;
}
bool same(int x, int y) {
return get(x) == get(y);
}
bool F(int x) { // 判断连通块内是否存在自环
return f[get(x)];
}
int size(int x) { // 输出连通块中点的数量
return p[get(x)];
}
int E(int x) { // 输出连通块中边的数量
return e[get(x)];
}
};
void solve() {
int n, m;
cin >> n >> m;
vector<int> in(n + 1);
DSU dsu(n);
for (int i = 0; i < m; i++) {
int u, v;
cin >> u >> v;
dsu.merge(u, v);
in[u]++, in[v]++;
}
int siz = 0, ans = 0;
for (int i = 1; i <= n; i++) {
if (dsu.get(i) == i) {
siz++;
int t = dsu.size(i);
if (t % 2) ans += t;
else ans += t - 1;
}
}
// if ((siz + 1) % 2 == 0) ans += m;
// for (int i = 1; i <= n; i++) {
// if ((siz - in[i]) % 2 == 0) ans++;
// }
cout << ans << '\n';
}
int main() {
// ios::sync_with_stdio(0); cin.tie(0), cout.tie(0);
int t = 1;
// cin >> t;
while(t--) solve();
return 0;
}
/*
3 1
1 2
4 3
1 2
2 3
3 4
*/
Details
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Test #1:
score: 100
Accepted
time: 0ms
memory: 3548kb
input:
3 1 1 2
output:
2
result:
ok 1 number(s): "2"
Test #2:
score: 0
Accepted
time: 0ms
memory: 3616kb
input:
4 3 1 2 2 3 3 4
output:
3
result:
ok 1 number(s): "3"
Test #3:
score: -100
Wrong Answer
time: 1ms
memory: 5304kb
input:
100000 1 4647 17816
output:
99999
result:
wrong answer 1st numbers differ - expected: '1', found: '99999'