【LG3703】[SDOI2017]树点涂色
【LG3703】[SDOI2017]树点涂色
题面
题解
一次只能染根到\(x\)，且染的颜色未出现过
这句话是我们解题的关键。
设\(x\)到根的颜色数为\(f(x)\)，则\(u\)到\(v\)的颜色数：\(f(u)+f(v)-f(lca_{u,v})+1\)
想一想，为什么？
很显然，如果没有\(1\)操作，我们直接树剖维护一下就可以了。
但是现在有了\(1\)操作。。。
这个\(1\)操作，其实是拉一条从\(x\)到根的链，染成一种颜色
这是不是很像\(LCT\)的\(access\)呢？
这样的话，我们就搞一颗\(LCT\)，\(access\)时，
因为每断一颗子树，那棵子树内必然要多加一个颜色段就是一个子树加，
每连上一颗子树，那棵子树内必然重复一个颜色段就是一个子树减。
那么我们用树剖维护每个点的\(f(x)\)
并魔改一下\(access\)即可
代码：
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
inline int gi() {
register int data = 0, w = 1;
register char ch = 0;
while (!isdigit(ch) && ch != '-') ch = getchar();
if (ch == '-') w = -1, ch = getchar();
while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar();
return w * data;
}
const int MAX_N = 1e5 + 5;
struct Graph { int to, next; } e[MAX_N << 1]; int fir[MAX_N], e_cnt;
void clearGraph() { memset(fir, -1, sizeof(fir)); e_cnt = 0; }
void Add_Edge(int u, int v) { e[e_cnt] = (Graph){v, fir[u]}, fir[u] = e_cnt++; }
int N, M;
int dfn[MAX_N], L[MAX_N], R[MAX_N], top[MAX_N];
int dep[MAX_N], fa[MAX_N], son[MAX_N], size[MAX_N], tim;
void dfs1(int x) {
dep[x] = dep[fa[x]] + 1, size[x] = 1;
for (int i = fir[x]; ~i; i = e[i].next) {
int v = e[i].to; if (v == fa[x]) continue;
fa[v] = x; dfs1(v); size[x] += size[v];
if (size[son[x]] < size[v]) son[x] = v;
}
}
void dfs2(int x, int tp) {
top[x] = tp, L[x] = ++tim, dfn[tim] = x;
if (son[x]) dfs2(son[x], tp);
for (int i = fir[x]; ~i; i = e[i].next) {
int v = e[i].to; if (v == fa[x] || v == son[x]) continue;
dfs2(v, v);
}
R[x] = tim;
}
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]];
}
return dep[x] < dep[y] ? x : y;
}
#define lson (o << 1)
#define rson (o << 1 | 1)
int val[MAX_N << 2], tag[MAX_N << 2];
void pushup(int o) { val[o] = max(val[lson], val[rson]); }
void puttag(int o, int v) { tag[o] += v; val[o] += v; }
void pushdown(int o, int l, int r) {
if (l == r || !tag[o]) return ;
puttag(lson, tag[o]);
puttag(rson, tag[o]);
tag[o] = 0;
}
void build(int o, int l, int r) {
if (l == r) return (void)(val[o] = dep[dfn[l]]);
int mid = (l + r) >> 1;
build(lson, l, mid), build(rson, mid + 1, r);
pushup(o);
}
void modify(int o, int l, int r, int ql, int qr, int v) {
if (ql <= l && r <= qr) return (void)(puttag(o, v));
pushdown(o, l, r);
int mid = (l + r) >> 1;
if (ql <= mid) modify(lson, l, mid, ql, qr, v);
if (qr > mid) modify(rson, mid + 1, r, ql, qr, v);
pushup(o);
}
int query(int o, int l, int r, int ql, int qr) {
pushdown(o, l, r);
if (ql <= l && r <= qr) return val[o];
int mid = (l + r) >> 1, res = 0;
if (ql <= mid) res = max(res, query(lson, l, mid, ql, qr));
if (qr > mid) res = max(res, query(rson, mid + 1, r, ql, qr));
return res;
}
struct Node { int ch[2], fa; bool rev; } t[MAX_N];
bool get(int x) { return t[t[x].fa].ch[1] == x; }
bool nroot(int x) { return t[t[x].fa].ch[0] == x || t[t[x].fa].ch[1] == x; }
void rotate(int x) {
int y = t[x].fa, z = t[y].fa, k = get(x);
if (nroot(y)) t[z].ch[get(y)] = x;
t[x].fa = z;
t[t[x].ch[k ^ 1]].fa = y, t[y].ch[k] = t[x].ch[k ^ 1];
t[y].fa = x, t[x].ch[k ^ 1] = y;
}
void splay(int x) {
while (nroot(x)) {
int y = t[x].fa;
if (nroot(y)) get(x) ^ get(y) ? rotate(x) : rotate(y);
rotate(x);
}
}
int findroot(int x) { while (t[x].ch[0]) x = t[x].ch[0]; return x; }
void access(int x) {
for (int y = 0; x; y = x, x = t[x].fa) {
splay(x);
if (t[x].ch[1]) {
int rt = findroot(t[x].ch[1]);
modify(1, 1, N, L[rt], R[rt], 1);
}
t[x].ch[1] = y;
if (t[x].ch[1]) {
int rt = findroot(t[x].ch[1]);
modify(1, 1, N, L[rt], R[rt], -1);
}
}
}
int main () {
clearGraph();
N = gi(), M = gi();
for (int i = 1; i < N; i++) {
int u = gi(), v = gi();
Add_Edge(u, v);
Add_Edge(v, u);
}
dfs1(1), dfs2(1, 1);
for (int x = 2; x <= N; x++) t[x].fa = fa[x];
build(1, 1, N);
while (M--) {
int op = gi();
if (op == 1) access(gi());
if (op == 2) {
int u = gi(), v = gi(), lca = LCA(u, v);
printf("%d\n", query(1, 1, N, L[u], L[u]) + query(1, 1, N, L[v], L[v])
- 2 * query(1, 1, N, L[lca], L[lca]) + 1);
}
if (op == 3) {
int x = gi();
printf("%d\n", query(1, 1, N, L[x], R[x]));
}
}
return 0;
}
