NC24141 [USACO 2011 Dec G]Grass Planting

题目链接

题目

题目描述

Farmer John has N barren pastures (2 <= N <= 100,000) connected by N-1 bidirectional roads, such that there is exactly one path between any two pastures. Bessie, a cow who loves her grazing time, often complains about how there is no grass on the roads between pastures. Farmer John loves Bessie very much, and today he is finally going to plant grass on the roads. He will do so using a procedure consisting of M steps (1 <= M <= 100,000).
At each step one of two things will happen:
- FJ will choose two pastures, and plant a patch of grass along each road in between the two pastures, or,
- Bessie will ask about how many patches of grass on a particular road, and Farmer John must answer her question.
Farmer John is a very poor counter -- help him answer Bessie's questions!

输入描述

• Line 1: Two space-separated integers N and M
• Lines 2..N: Two space-separated integers describing the endpoints of a road.
• Lines N+1..N+M: Line i+1 describes step i. The first character of the line is either P or Q, which describes whether or not FJ is planting grass or simply querying. This is followed by two space-separated integers Ai and Bi (1 <= Ai, Bi <= N) which describe FJ's action or query.

输出描述

• Lines 1..???: Each line has the answer to a query, appearing in the
  same order as the queries appear in the input.

示例1

输入

4 6
1 4
2 4
3 4
P 2 3
P 1 3
Q 3 4
P 1 4
Q 2 4
Q 1 4

输出

2
1
2

题解

知识点：树链剖分，线段树。

通常树链剖分维护的是点权，但这里需要维护边权，因此我们考虑将边权映射到点权上，考虑用边的下端点代替一条边（上端点会导致非一对一映射）。

需要注意，更新操作时，对最后一段 $(u,v)$ 更新时，$u$ 对应的那条边是不需要的，所以更新 $[L[u]+1,L[v]]$ 的点权。

时间复杂度 $O((n+m)\log n)$

空间复杂度 $O(n)$

代码

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
 
struct HLD {
    vector<int> siz, dep, fat, son, top, dfn, L, R;
 
    HLD() {}
    HLD(int rt, const vector<vector<int>> &g) { init(rt, g); }
 
    void init(int rt, const vector<vector<int>> &g) {
        assert(g.size() >= rt + 1);
        int n = g.size() - 1;
        siz.assign(n + 1, 0);
        dep.assign(n + 1, 0);
        fat.assign(n + 1, 0);
        son.assign(n + 1, 0);
        top.assign(n + 1, 0);
        dfn.assign(n + 1, 0);
        L.assign(n + 1, 0);
        R.assign(n + 1, 0);
 
        function<void(int, int)> dfsA = [&](int u, int fa) {
            siz[u] = 1;
            dep[u] = dep[fa] + 1;
            fat[u] = fa;
            for (auto v : g[u]) {
                if (v == fa) continue;
                dfsA(v, u);
                siz[u] += siz[v];
                if (siz[v] > siz[son[u]]) son[u] = v;
            }
        };
        dfsA(rt, 0);
 
        int dfncnt = 0;
        function<void(int, int)> dfsB = [&](int u, int tp) {
            top[u] = tp;
            dfn[++dfncnt] = u;
            L[u] = dfncnt;
            if (son[u]) dfsB(son[u], tp);
            for (auto v : g[u]) {
                if (v == fat[u] || v == son[u]) continue;
                dfsB(v, v);
            }
            R[u] = dfncnt;
        };
        dfsB(rt, rt);
    }
};
 
template <class T>
class Fenwick {
    int n;
    vector<T> node;
 
public:
    Fenwick(int _n = 0) { init(_n); }
 
    void init(int _n) {
        n = _n;
        node.assign(n + 1, T());
    }
 
    void update(int x, T val) { for (int i = x;i <= n;i += i & -i) node[i] += val; }
 
    T query(int x) {
        T ans = T();
        for (int i = x;i >= 1;i -= i & -i) ans += node[i];
        return ans;
    }
};
 
struct T {
    int sum = 0;
    T &operator+=(const T &x) { return sum += x.sum, *this; }
};
 
const int N = 1e5 + 7;
vector<int> g[N];
HLD hld;
Fenwick<T> fw;
 
void path_update(int u, int v) {
    auto &top = hld.top;
    auto &dep = hld.dep;
    auto &fat = hld.fat;
    auto &L = hld.L;
    while (top[u] != top[v]) {
        if (dep[top[u]] < dep[top[v]]) swap(u, v);
        fw.update(L[top[u]], { 1 });
        fw.update(L[u] + 1, { -1 });
        u = fat[top[u]];
    }
    if (dep[u] > dep[v]) swap(u, v);
    fw.update(L[u] + 1, { 1 });
    fw.update(L[v] + 1, { -1 });
}
 
int main() {
    std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n, m;
    cin >> n >> m;
    for (int i = 1;i <= n - 1;i++) {
        int u, v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    hld.init(1, vector<vector<int>>(g, g + n + 1));
 
    fw.init(n);
 
    while (m--) {
        char op;
        cin >> op;
        if (op == 'P') {
            int u, v;
            cin >> u >> v;
            path_update(u, v);
        }
        else {
            int u, v;
            cin >> u >> v;
            if (hld.dep[u] < hld.dep[v]) swap(u, v);
            cout << fw.query(hld.L[u]).sum << '\n';
        }
    }
    return 0;
}