[CF342E] Xenia and Tree - 分块,ST表,LCA
Description
给定一颗 (n) 个结点的树,初始时 (1) 号结点为红色,其余为蓝色。要求支持将一个结点变为红色,或者询问一个点到最近的红色点的距离。
Solution
对整个操作序列分块,则所有的影响分为块间的和块内的。块内的用 ST 表 LCA 来处理,块间的预处理好,即每处理完一个块,我们就用 BFS 计算好这个块的后续影响。复杂度 (O(n sqrt n))
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1000005;
int n, m;
vector<int> g[N];
namespace _lca
{
int rt, st[N][20], dep[N], dis[N], vis[N], ind, lg2[N], s[N], bg[N], ed[N];
void dfs(int p)
{
vis[p] = 1;
s[++ind] = p;
bg[p] = ind;
for (int q : g[p])
{
if (vis[q] == 0)
{
dep[q] = dep[p] + 1;
dfs(q);
s[++ind] = p;
}
}
ed[p] = ind;
}
int lca(int p, int q)
{
if (p == q)
return p;
p = bg[p];
q = bg[q];
if (p > q)
swap(p, q);
int l = lg2[q - p + 1];
int x = st[p][l];
int y = st[q - (1 << l) + 1][l];
return s[dis[x] < dis[y] ? x : y];
}
int dist(int p, int q)
{
int l = lca(p, q);
return dep[p] + dep[q] - 2 * dep[l];
}
void solve()
{
rt = 1;
for (int i = 0; i <= 18; i++)
for (int j = 1 << i; j < 1 << (i + 1); j++)
lg2[j] = i;
dfs(rt);
for (int i = 1; i <= ind; i++)
dis[i] = dep[s[i]];
for (int i = 1; i <= ind; i++)
st[i][0] = i;
for (int i = 1; i <= 17; i++)
{
for (int j = 1; j <= ind; j++)
{
st[j][i] = dis[st[j][i - 1]] < dis[st[j + (1 << (i - 1))][i - 1]] ? st[j][i - 1] : st[j + (1 << (i - 1))][i - 1];
}
}
}
} // namespace _lca
int get_dist(int p, int q)
{
return _lca::dist(p, q);
}
signed main()
{
ios::sync_with_stdio(false);
cin >> n >> m;
for (int i = 1; i < n; i++)
{
int t1, t2;
cin >> t1 >> t2;
g[t1].push_back(t2);
g[t2].push_back(t1);
}
_lca::solve();
int b = sqrt(m);
vector<int> buf;
vector<int> dis(n + 2);
for (int i = 1; i <= n; i++)
{
dis[i] = 1e18;
}
buf.push_back(1);
for (int i = 1; i <= m; i++)
{
int type, v;
cin >> type >> v;
if (type == 1)
{
buf.push_back(v);
}
else
{
int ans = dis[v];
for (auto p : buf)
{
ans = min(ans, get_dist(p, v));
}
cout << ans << endl;
}
if (i % b == 0)
{
queue<int> que;
for (auto p : buf)
{
que.push(p);
dis[p] = 0;
}
buf.clear();
while (que.size())
{
int p = que.front();
que.pop();
for (auto q : g[p])
{
if (dis[q] > dis[p] + 1)
{
dis[q] = dis[p] + 1;
que.push(q);
}
}
}
}
}
}