1136 - Counting Paths
28 Jan 2021 — Tags: None
Click to show code.
using namespace std;
using ll = long long;
using ii = pair<int, int>;
using vi = vector<int>;
template <typename InputIterator,
typename T = typename iterator_traits<InputIterator>::value_type>
void read_n(InputIterator it, int n)
{
copy_n(istream_iterator<T>(cin), n, it);
}
template <typename InputIterator,
typename T = typename iterator_traits<InputIterator>::value_type>
void write(InputIterator first, InputIterator last, const char *delim = "\n")
{
copy(first, last, ostream_iterator<T>(cout, delim));
}
int const NMAX = 2e5 + 11;
int const LMAX = 20;
int n, timer, val[NMAX], tin[NMAX], tout[NMAX], up[NMAX][LMAX], ans[NMAX];
vi g[NMAX];
void preprocess(int u, int p)
{
tin[u] = ++timer;
up[u][0] = p;
for (int i = 1; i < LMAX; ++i)
up[u][i] = up[up[u][i - 1]][i - 1];
for (auto v : g[u])
if (v != p)
preprocess(v, u);
tout[u] = ++timer;
}
void solve(int u, int p)
{
for (auto v : g[u])
{
if (v == p)
continue;
solve(v, u);
val[u] += val[v];
}
ans[u] = val[u];
}
bool is_ancestor(int u, int v)
{
return tin[u] <= tin[v] && tout[u] >= tout[v];
}
int lca(int u, int v)
{
if (is_ancestor(u, v))
return u;
if (is_ancestor(v, u))
return v;
for (int i = LMAX - 1; i >= 0; --i)
if (!is_ancestor(up[u][i], v))
u = up[u][i];
return up[u][0];
}
int main(void)
{
ios::sync_with_stdio(false), cin.tie(NULL);
int m;
cin >> n >> m;
for (int i = 0; i < n - 1; ++i)
{
int u, v;
cin >> u >> v;
g[u].push_back(v);
g[v].push_back(u);
}
int root = 1;
preprocess(root, root);
while (m--)
{
int u, v, l;
cin >> u >> v;
val[u] += 1;
val[v] += 1;
l = lca(u, v);
val[l] -= 1;
if (l != root)
val[up[l][0]] -= 1;
}
solve(1, 1);
write(ans + 1, ans + n + 1, " "), cout << endl;
return 0;
}