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QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#115879 | #5307. Subgraph Isomorphism | PetroTarnavskyi# | WA | 37ms | 18964kb | C++17 | 2.2kb | 2023-06-27 17:11:04 | 2023-06-27 17:11:06 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
#define SZ(a) (int)a.size()
#define ALL(a) a.begin(), a.end()
#define FOR(i, a, b) for (int i = (a); i<(b); ++i)
#define RFOR(i, b, a) for (int i = (b)-1; i>=(a); --i)
#define MP make_pair
#define PB push_back
#define F first
#define S second
typedef long long LL;
typedef pair<int, int> PII;
typedef vector<int> VI;
const int N = 400447;
const int mod[2] = {int(1e9) + 9, int(1e9) + 7};
const int p = 47;
VI g[N];
int pw[N][2];
int par[N];
bool used[N];
VI cycl;
int add(int a, int b, int md)
{
a += b;
if (a >= md)
a -= md;
return a;
}
int mult(int a, int b, int md)
{
return LL(a) * b % md;
}
bool dfs1(int v, int from = -1)
{
par[v] = from;
used[v] = 1;
bool ok = false;
for (auto to : g[v])
{
if (to == from) continue;
if (used[to])
{
cycl.PB(to);
while (v != to)
{
cycl.PB(v);
v = par[v];
}
return true;
}
else
ok |= dfs1(to, v);
if (ok) return ok;
}
return false;
}
int sz[N];
PII dfs2(int v)
{
sz[v] = 1;
used[v] = 1;
vector<PII> vec;
for (auto to : g[v])
if (!used[to]){
vec.PB(dfs2(to));
sz[v] += sz[to];
}
sort(ALL(vec));
PII hs = {(sz[v] + (sz[v] ^ 7)) , (sz[v] + (sz[v] ^ 7))};
FOR (i, 0, SZ(vec)) hs = { add(hs.first, mult(pw[i + 1][0], vec[i].first, mod[0]), mod[0]),
add(hs.second, mult(pw[i + 1][1], vec[i].second, mod[1]), mod[1])};
return hs;
}
void solve()
{
int n, m;
cin >> n >> m;
FOR (i, 0, m)
{
int a, b;
cin >> a >> b;
a--, b--;
g[a].PB(b);
g[b].PB(a);
}
if (m == n - 1)
{
cout << "YES\n";
}
else if (m > n)
{
cout << "NO\n";
}
else
{
dfs1(0, -1);
FOR (i, 0, n) used[i] = 0;
set<PII> s;
for (auto v : cycl) used[v] = 1;
for (auto v : cycl) s.insert(dfs2(v));
cout << (SZ(s) < 2 ? "YES\n" : "NO\n");
}
cycl.clear();
FOR (i, 0, n)
{
g[i].clear();
used[i] = 0;
}
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
pw[0][0] = pw[0][1] = 1;
FOR (i, 1, N) FOR (j, 0, 2) pw[i][j] = mult(pw[i - 1][j], p, mod[j]);
int t;
cin >> t;
while (t--)
{
solve();
}
return 0;
}
详细
Test #1:
score: 100
Accepted
time: 1ms
memory: 17980kb
input:
4 7 6 1 2 2 3 3 4 4 5 5 6 3 7 3 3 1 2 2 3 3 1 5 5 1 2 2 3 3 4 4 1 1 5 1 0
output:
YES YES NO YES
result:
ok 4 token(s): yes count is 3, no count is 1
Test #2:
score: -100
Wrong Answer
time: 37ms
memory: 18964kb
input:
33192 2 1 1 2 3 2 1 3 2 3 3 3 1 2 1 3 2 3 4 3 1 4 2 4 3 4 4 3 1 3 1 4 2 4 4 4 1 3 1 4 2 4 3 4 4 4 1 3 1 4 2 3 2 4 4 5 1 3 1 4 2 3 2 4 3 4 4 6 1 2 1 3 1 4 2 3 2 4 3 4 5 4 1 5 2 5 3 5 4 5 5 4 1 4 1 5 2 5 3 5 5 5 1 4 1 5 2 5 3 5 4 5 5 5 1 4 1 5 2 4 3 5 4 5 5 5 1 4 1 5 2 4 2 5 3 5 5 6 1 4 1 5 2 4 2 5 3 ...
output:
YES YES YES YES YES NO YES NO NO YES YES NO NO NO NO NO NO YES NO NO NO NO YES NO NO NO NO NO NO NO YES YES NO YES YES NO NO NO NO NO NO NO NO NO YES NO NO NO YES NO NO NO NO NO NO NO NO NO NO YES NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO YES NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO NO ...
result:
wrong answer expected YES, found NO [39th token]