题意
题目链接
Sol
下面的代码是(O(nlog^3n))的暴力。
因为从一个点向上只会跳(logn)次，所以可以暴力的把未经过的处理出来然后每个点开个multiset维护最大值
代码语言：javascript复制#include<bits/stdc .h>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long
#define LL long long
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 2e5 10, SS = MAXN * 4, mod = 1e9 7, INF = 1e9 10;
const double eps = 1e-9;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline LL add(A x, B y) {if(x y < 0) return x y mod; return x y >= mod ? x y - mod : x y;}
template <typename A, typename B> inline void add2(A &x, B y) {if(x y < 0) x = x y mod; else x = (x y >= mod ? x y - mod : x y);}
template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod mod) % mod;}
template <typename A> inline void debug(A a){cout << a << 'n';}
template <typename A> inline LL sqr(A x){return 1ll * x * x;}
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 c - '0', c = getchar();
return x * f;
}
int N, Q, fa[MAXN], siz[MAXN], son[MAXN], id[MAXN], top[MAXN], dep[MAXN], times;
vector<int> v[MAXN];
void dfs1(int x, int _fa) {
siz[x] = 1; dep[x] = dep[_fa] 1; fa[x] = _fa;
for(auto &to : v[x]) {
if(to == _fa) continue;
dfs1(to, x);
siz[x] = siz[to];
if(siz[to] > siz[son[x]]) son[x] = to;
}
}
void dfs2(int x, int topf) {
top[x] = topf; id[x] = times;
if(!son[x]) return ;
dfs2(son[x], topf);
for(auto &to : v[x]) {
if(top[to]) continue;
dfs2(to, to);
}
}
multiset<int> s[SS];
struct Query {
int a, b, v;
}q[MAXN];
vector<Pair> line[MAXN];
int ls[SS], rs[SS], root, tot;
void Erase(multiset<int> &s, int v) {
auto it = s.find(v);
if(it != s.end()) s.erase(it);
}
void Get(vector<Pair> &v, int x, int y) {
vector<Pair> tmp;
while(top[x] ^ top[y]) {
if(dep[top[x]] < dep[top[y]]) swap(x, y);
tmp.push_back({id[top[x]], id[x]});
x = fa[top[x]];
}
if(dep[x] > dep[y]) swap(x, y);
tmp.push_back({id[x], id[y]});
sort(tmp.begin(), tmp.end());
int las = 1;
for(auto x : tmp) {
if(las <= x.fi - 1) v.push_back({las, x.fi - 1});
las = x.se 1;
}
if(las <= N) v.push_back({las, N});
}
int Mx(multiset<int> &s) {
if(s.empty()) return -1;
auto it = s.end(); it--;
return *it;
}
void IntAdd(int &k, int l, int r, int ql, int qr, int v, int opt) {
if(!k) k = tot;
if(ql <= l && r <= qr) {
if(opt == 1) s[k].insert(v);
else Erase(s[k], v);
return ;
}
int mid = l r >> 1;
if(ql <= mid) IntAdd(ls[k], l, mid, ql, qr, v, opt);
if(qr > mid) IntAdd(rs[k], mid 1, r, ql, qr, v, opt);
}
int Query(int k, int l, int r, int p) {
if(!k) return -1;
int ans = Mx(s[k]), mid = l r >> 1;
if(l == r) return Mx(s[k]);
if(p <= mid) chmax(ans, Query(ls[k], l, mid, p));
else chmax(ans, Query(rs[k], mid 1, r, p));
return ans;
}
void TreeAdd(int x, int y, int v, int opt) {
while(top[x] ^ top[y]) {
if(dep[top[x]] < dep[top[y]]) swap(x, y);
IntAdd(root, 1, N, id[top[x]], id[x], v, opt);
x = fa[top[x]];
}
if(dep[x] > dep[y]) swap(x, y);
IntAdd(root, 1, N, id[x], id[y], v, opt);
}
void Add(int ti, int opt) {
int x = q[ti].a, y = q[ti].b, v = q[ti].v;
if(opt == 1) Get(line[ti], x, y);
for(auto x : line[ti])
IntAdd(root, 1, N, x.fi, x.se, v, opt);
}
signed main() {
// Fin(a); Fout(b);
N = read(); Q = read();
for(int i = 1; i <= N - 1; i ) {
int x = read(), y = read();
v[x].push_back(y);
v[y].push_back(x);
}
dfs1(1, 0);
dfs2(1, 1);
for(int i = 1; i <= Q; i ) {
int opt = read();
if(opt == 0) {
int a = read(), b = read(), v = read(); q[i] = {a, b, v};
Add(i, 1);
} else if(opt == 1) {
int ti = read();
Add(ti, -1);
} else if(opt == 2) {
int x = read();
printf("%dn", Query(root, 1, N, id[x]));
}
}
return 0;
}
