题意
给定一棵树，每次操作将一个点染成黑色。
求询问的点到所有黑点的路径编号最小值。
** 数据保证第一次为染色操作 **
Sol
注意到保证第一次为染色。
考虑钦定根节点为染色的点。
那么对于所有染色操作，暴力记录染色的点到根节点的路径上所有点的贡献。
每个点只会贡献一次，这部分是 \(O(n)\)。
考虑询问的点到根节点的贡献。预处理即可 \(O (n + q)\)。
Code
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <array>
#include <bitset>
using namespace std;
#ifdef ONLINE_JUDGE
#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
char buf[1 << 23], *p1 = buf, *p2 = buf, ubuf[1 << 23], *u = ubuf;
#endif
int read() {
int p = 0, flg = 1;
char c = getchar();
while (c < '0' || c > '9') {
if (c == '-') flg = -1;
c = getchar();
}
while (c >= '0' && c <= '9') {
p = p * 10 + c - '0';
c = getchar();
}
return p * flg;
}
string read_() {
string ans;
char c = getchar();
while (c != 'J' && c != 'Q')
c = getchar();
while (c == 'J' || c == 'Q') {
ans += c;
c = getchar();
}
return ans;
}
void write(int x) {
if (x < 0) {
x = -x;
putchar('-');
}
if (x > 9) {
write(x / 10);
}
putchar(x % 10 + '0');
}
const int N = 1e6 + 5, M = 2e6 + 5;
namespace G {
array <int, N> fir;
array <int, M> nex, to;
int cnt;
void add(int x, int y) {
cnt++;
nex[cnt] = fir[x];
to[cnt] = y;
fir[x] = cnt;
}
}
/*
namespace Hpt {
using G::fir, G::nex, G::to;
array <int, N> fa, siz, son, dep;
void dfs1(int x) {
siz[x] = 1;
for (int i = fir[x]; i; i = nex[i]) {
if (to[i] == fa[x]) continue;
fa[to[i]] = x;
dep[to[i]] = dep[x] + 1;
dfs1(to[i]);
if (siz[to[i]] > siz[son[x]]) son[x] = to[i];
}
}
array <int, N> dfn, idx, top;
int cnt;
void dfs2(int x, int Mgn) {
cnt++;
dfn[x] = cnt;
idx[cnt] = x;
top[x] = Mgn;
if (son[x]) dfs2(son[x], Mgn);
for (int i = fir[x]; i; i = nex[i]) {
if (to[i] == son[x] || to[i] == fa[x]) continue;
dfs2(to[i], to[i]);
}
}
int lca(int x, int y) {
while (top[x] != top[y]) {
if (dep[top[x]] < dep[top[y]]) swap(x, y);
x = fa[top[x]];
}
if (dfn[x] > dfn[y]) swap(x, y);
return x;
}
}
*/
using G::fir, G::nex, G::to;
bitset <N> vis;
array <int, N> fa, dis;
int dfs(int x, int rt) {
int ans = 0x7f7f7f7f;
while (x != rt && !vis[x]) {
ans = min(x, ans);
vis[x] = 1;
x = fa[x];
}
return ans;
}
void _dfs(int x) {
dis[x] = min(dis[x], x);
for (int i = fir[x]; i; i = nex[i]) {
if (to[i] == fa[x]) continue;
dis[to[i]] = min(dis[to[i]], dis[x]);
fa[to[i]] = x;
_dfs(to[i]);
}
}
int main() {
freopen("network.in", "r", stdin);
freopen("network.out", "w", stdout);
int n = read(), q = read() - 1;
for (int i = 2; i <= n; i++) {
int x = read(), y = read();
G::add(x, y), G::add(y, x);
}
dis.fill(0x7f7f7f7f);
int lcaST = 0x7f7f7f7f;
int rt = read();
_dfs(rt);
/* Hpt::dfs1(rt), Hpt::dfs2(rt, 0); */
while (q--) {
string opt = read_();
if (opt[0] == 'J') {
int x = read();
lcaST = min(lcaST, dfs(x, rt));
}
else {
int x = read();
write(min(lcaST, dis[x])), puts("");
}
}
return 0;
}