loj#2049. 「HNOI2016」网络(set 树剖 暴力)

题意

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

#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("%d\n", query(root, 1, n, id[x]));
        }
    }
    return 0;
}