https://www.luogu.org/problemnew/show/U18430

思路比较好想

树链剖分

对于1操作

  只需将以该点为根的子树打标记,将所有数存入数组排序

  每次进行1操作时,判断若该点已经被标记过,则操作无效

对于操作2

  在原树中求出lca,若lca被标记过,lca就是询问的两个点中

  编号较小的那个,lca值不改变

对于操作3

  若该点打过标记,输出该点与根的这条链上被打过标记的

  深度最小节点编号a与该节点编号的差+a的深度

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath> using namespace std;
const int N = 1e6 + ; #define yxy getchar()
#define lson jd << 1
#define rson jd << 1 | 1 int n, Q, now = , tim, w1, w2, w3;
int head[N], Num[N], deep[N], fa[N], lnum[N], rnum[N], son[N], topp[N], siz[N], tree[N], g_min[N << ];
struct Node{int u, v, nxt;} G[N << ]; inline int read(){
int x = ; char c = yxy;
while(c < '' || c > '') c = yxy;
while(c >= '' && c <= '') x = x * + c - '', c = yxy;
return x;
} inline void add(int u, int v){
G[now].u = u; G[now].v = v; G[now].nxt = head[u]; head[u] = now ++;
} void dfs_1(int u, int father, int dep){
fa[u] = father;
deep[u] = dep;
siz[u] = ;
for(int i = head[u]; ~ i; i = G[i].nxt){
int v = G[i].v;
if(v != father){
dfs_1(v, u, dep + );
siz[u] += siz[v];
if(siz[v] > siz[son[u]]) son[u] = v;
}
}
} void dfs_2(int u, int tp){
topp[u] = tp;
lnum[u] = ++ tim;
tree[u] = tim;
Num[tim] = u;
if(!son[u]) {rnum[u] = tim; return ;}
dfs_2(son[u], tp);
for(int i = head[u]; ~ i; i = G[i].nxt){
int v = G[i].v;
if(v != fa[u] && v != son[u]) dfs_2(v, v);
}
rnum[u] = tim;
} void Poi_A(int l, int r, int jd, int x){
if(l == r) {
if(g_min[jd]) w1 = ;
return ;
}
if(g_min[jd]) {
g_min[lson] = g_min[jd];
g_min[rson] = g_min[jd];
}
int mid = (l + r) >> ;
if(x <= mid) Poi_A(l, mid, lson, x);
else Poi_A(mid + , r, rson, x);
} void Sec_G(int l, int r, int jd, int x, int y, int yg){
if(x <= l && r <= y){
g_min[jd] = yg;
return ;
}
int mid = (l + r) >> ;
if(x <= mid) Sec_G(l, mid, lson, x, y, yg);
if(y > mid) Sec_G(mid + , r, rson, x, y, yg);
} int calc_lca(int x, int y){
int tp1 = topp[x], tp2 = topp[y];
while(tp1 != tp2){
if(deep[tp1] < deep[tp2]) swap(tp1, tp2), swap(x, y);
x = fa[tp1];
tp1 = topp[x];
}
return deep[x] > deep[y] ? y : x;
} void Sec_A(int l, int r, int jd, int x, int y){
if(x <= l && r <= y){
w2 = g_min[jd];
return ;
}
if(g_min[jd]) {
g_min[lson] = g_min[jd];
g_min[rson] = g_min[jd];
}
int mid = (l + r) >> ;
if(x <= mid) Sec_A(l, mid, lson, x, y);
if(y > mid) Sec_A(mid + , r, rson, x, y);
} int get_answer(int l, int r, int x){
int L = l, R = r, ret;
while(L <= R){
int mid = (L + R) >> ;
if(Num[mid] <= x) L = mid + , ret = mid;
else R = mid - ;
}
return ret - l;
} int main()
{
n = read(); Q = read();
for(int i = ; i <= n; i ++) head[i] = -;
for(int i = ; i < n; i ++) {
int u = read(), v = read();
add(u, v); add(v, u);
}
dfs_1(, , );
dfs_2(, );
while(Q --){
int opt = read();
if(opt == ){
int x = read();
if(!son[x]) continue ;
w1 = ;
Poi_A(, n, , tree[x]);
if(w1) continue ;
Sec_G(, n, , lnum[x], rnum[x], x);
sort(Num + lnum[x] + , Num + rnum[x] + );
}
else if(opt == ){
int x = read(), y = read();
int Lca_now = calc_lca(x, y);
w1 = ;
Poi_A(, n, , tree[Lca_now]);
if(w1) printf("%d\n", min(x, y));
else printf("%d\n", Lca_now);
}
else {
int x = read();
w1 = ;
Poi_A(, n, , tree[x]);
if(!w1) printf("%d\n", deep[x]);
else {
Sec_A(, n, , lnum[x], rnum[x]);
printf("%d\n", get_answer(lnum[w2], rnum[w2], x) + deep[w2]);
}
}
}
return ;
}

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