中文题意

给你个节点的一棵树，并且存在次操作，每次操作把节点之间路径上全部的点权值加一。问操作次操作后整棵树最大的权值是多少？

Solution

读懂题的话，就是树上差分模板题，我实现的办法是树链剖分，会了之后就觉得比倍增快乐多了。给个OIWiki的差分链接

树上点差分就是：

#include <bits/stdc++.h>
using namespace std;
#define rep(i, sta, en) for(int i=sta; i<=en; ++i)
#define repp(i, sta, en) for(int i=sta; i>=en; --i)
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void print(ll x, int op = 10) { if (!x) { putchar('0'); if (op)    putchar(op); return; }    char F[40]; ll tmp = x > 0 ? x : -x;    if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]);    if (op)    putchar(op); }

const int N = 5e4 + 7;
int n, m;
int fa[N], dep[N], sz[N], son[N], top[N];
vector<int> edge[N];

void dfs1(int u) {
    son[u] = -1;
    sz[u] = 1;
    for (auto& v : edge[u]) {
        if (dep[v])    continue;
        dep[v] = dep[u] + 1;
        fa[v] = u;
        dfs1(v);
        sz[u] += sz[v];
        if (son[u] == -1 or sz[v] > sz[son[u]])
            son[u] = v;
    }
}

void dfs2(int u, int t) {
    top[u] = t;
    if (son[u] == -1)    return;
    dfs2(son[u], t);
    for (auto& v : edge[u])
        if (v != son[u] and v != fa[u])    dfs2(v, v);
}

int lca(int u, int v) {
    while (top[u] != top[v]) {
        if (dep[top[u]] > dep[top[v]])
            u = fa[top[u]];
        else
            v = fa[top[v]];
    }
    return dep[u] > dep[v] ? v : u;
}

int f[N], maxi;

void dfs3(int u, int fa) {
    for (auto& v : edge[u]) {
        if (v == fa)    continue;
        dfs3(v, u);
        f[u] += f[v];
    }
    maxi = max(maxi, f[u]);
}

void solve() {
    n = read(), m = read();
    rep(i, 2, n) {
        int u = read(), v = read();
        edge[u].push_back(v);
        edge[v].push_back(u);
    }
    dep[1] = 1;
    dfs1(1);
    dfs2(1, 1);
    while (m--) {
        int u = read(), v = read(), tmp = lca(u, v);
        ++f[u], --f[tmp];
        ++f[v], --f[fa[tmp]];
    }
    dfs3(1, 0);
    print(maxi);
}

int main() {
    //int T = read();    rep(_, 1, T)
    {
        solve();
    }
    return 0;
}