标签:upd query 子节点 return spoj 方向 namespace name using
https://www.spoj.com/problems/QTREE6/
考虑对0颜色和1颜色分别维护定根(no reverse)lct。
即在\(c[x]\)的lct上给\(x\)到\(fa[x]\)连一条边。
修改直接link、cut。
查询x的话,考虑access走到最上面的点,这个点实际上和x是不连通的,但是它的子节点(x方向的)的子树就是联通块大小。
所以lct要顺便维护子树和。
Code:
#include<bits/stdc++.h>
#define fo(i, x, y) for(int i = x, _b = y; i <= _b; i ++)
#define ff(i, x, y) for(int i = x, _b = y; i < _b; i ++)
#define fd(i, x, y) for(int i = x, _b = y; i >= _b; i --)
#define ll long long
#define pp printf
#define hh pp("\n")
using namespace std;
const int N = 1e5 + 5;
int n, x, y;
vector<int> e[N];
#define pb push_back
#define si size()
int fa[N];
void dg(int x) {
ff(_y, 0, e[x].si) {
int y = e[x][_y];
if(y == fa[x]) continue;
fa[y] = x;
dg(y);
}
}
int id[N][2], td, id2[N];
struct lct {
#define x0 t[x][0]
#define x1 t[x][1]
int fa[N], pf[N], t[N][2], s[N], ps[N];
int lr(int x) { return t[fa[x]][1] == x;}
void upd(int x) {
if(x) s[x] = 1 + ps[x] + s[x0] + s[x1];
}
void ro(int x) {
int y = fa[x], k = lr(x);
t[y][k] = t[x][!k]; if(t[x][!k]) fa[t[x][!k]] = y;
fa[x] = fa[y]; if(fa[y]) t[fa[y]][lr(y)] = x;
fa[y] = x, t[x][!k] = y, pf[x] = pf[y];
upd(y); upd(x);
}
void sp(int x, int y) {
for(; fa[x] != y; ro(x)) if(fa[fa[x]] != y)
ro(lr(x) == lr(fa[x]) ? fa[x] : x);
}
void ac(int x) {
int xx = x;
for(int y = 0; x; ) {
sp(x, 0), fa[x1] = 0, pf[x1] = x;
ps[x] += s[x1];
x1 = y, fa[y] = x, pf[y] = 0;
ps[x] -= s[y];
upd(x), y = x, x = pf[x];
}
sp(xx, 0);
}
void link(int x, int y) {
ac(y); ac(x);
pf[x] = y; ps[y] += s[x];
upd(y);
ac(x);
}
void cut(int x, int y) {
ac(x); sp(y, 0);
t[y][1] = fa[x] = pf[x] = 0;
upd(y);
}
int fl(int x) { return x0 ? fl(x0) : x;}
int qry(int x) {
ac(x);
int y = fl(x);
sp(y, 0);
return s[t[y][1]];
}
} l[2];
int c[N];
int m, op;
int main() {
scanf("%d", &n);
fo(i, 1, n - 1) {
scanf("%d %d", &x, &y);
e[x].pb(y); e[y].pb(x);
}
dg(1);
fa[1] = n + 1;
fo(i, 1, n + 1) fo(j, 0, 1)
l[j].s[i] = 1;
fo(i, 1, n) {
c[i] = 1;
l[1].link(i, fa[i]);
}
scanf("%d", &m);
fo(i, 1, m) {
scanf("%d %d", &op, &x);
if(op == 1) {
l[c[x]].cut(x, fa[x]);
c[x] = !c[x];
l[c[x]].link(x, fa[x]);
} else {
pp("%d\n", l[c[x]].qry(x));
}
}
}
SPOJ QTREE6 - Query on a tree VI(lct)
标签:upd query 子节点 return spoj 方向 namespace name using
原文地址:https://www.cnblogs.com/coldchair/p/12919594.html