原题链接戳这儿

SOLUTION

考虑一种非常\(naive\)的统计方法，就是对于每一个点\(u\)，我们维护它能到达的点集\(S_u\)，最后答案就是\(\frac{\sum\limits_{i=1}^{n}|S_i|}{2}\)

也就是说我们可以先树剖一下，对于每一个点都开一棵线段树，每次修改\(O(nlogn)\)地更新一下路径上的线段树，最后查询一下就行了

但是这样的复杂度是\(O(n^2log^2n)\)的，显然会炸。注意到每次是对一条链上的所有点操作，所以我们可以查分。又因为差分之后要把子树的贡献传上去，再上个线段树合并就行了，复杂度降为\(O(nlog^2n)\)

代码也比较好写，细节不多

#include <bits/stdc++.h>

using namespace std;

#define N 100000

#define ll long long

#define mp make_pair

#define pii pair<int, int>

#define pb push_back

#define mid ((l + r) >> 1)

int n, m;

vector<int> G[N + 5];

int sz[N + 5], fa[N + 5], d[N + 5], hson[N + 5], top[N + 5], dfn[N + 5], dfn_clk, id[N + 5];

int nid, root[N + 5], sumv[N << 7], ch[2][N << 7], addv[N << 7];

vector<pii> cf[N + 5], segs[N + 5];

ll ans;

void dfs1(int u, int pa) {

    sz[u] = 1;

    fa[u] = pa;

    for (int i = 0, v; i < G[u].size(); ++i) {

        v = G[u][i];

        if (v == pa) continue;

        d[v] = d[u] + 1;

        dfs1(v, u);

        sz[u] += sz[v];

        if (sz[v] > sz[hson[u]]) hson[u] = v;

    }

}

void dfs2(int u, int tp) {

    dfn[u] = ++dfn_clk;

    id[dfn_clk] = u;

    top[u] = tp;

    if (hson[u]) dfs2(hson[u], tp);

    for (int i = 0, v; i < G[u].size(); ++i) {

        v = G[u][i];

        if (v == fa[u] || v == hson[u]) continue;

        dfs2(v, v);

    }

}

int lca(int x, int y) {

    while (top[x] != top[y]) d[top[x]] > d[top[y]] ? x = fa[top[x]] : y = fa[top[y]];

    return d[x] < d[y] ? x : y;

}

void addModify(int x, int s, int t) {

    int z = lca(s, t);

    cf[s].pb(mp(x, 1)), cf[t].pb(mp(x, 1));

    cf[z].pb(mp(x, -1)), cf[fa[z]].pb(mp(x, -1));

    while (top[s] != top[z]) segs[x].pb(mp(dfn[top[s]], dfn[s])), s = fa[top[s]];

    segs[x].pb(mp(dfn[z], dfn[s]));

    while (top[t] != top[z]) segs[x].pb(mp(dfn[top[t]], dfn[t])), t = fa[top[t]];

    if (dfn[z] + 1 <= dfn[t]) segs[x].pb(mp(dfn[z] + 1, dfn[t]));

}

void pushup(int o, int l, int r) {

    if (addv[o]) sumv[o] = r - l + 1;

    else sumv[o] = sumv[ch[0][o]] + sumv[ch[1][o]];

}

void add(int &o, int l, int r, int L, int R, int k) {

    if (!o) o = ++nid;

    if (L <= l && r <= R) {

        addv[o] += k;

        pushup(o, l, r);

        return ;

    }

    if (L <= mid) add(ch[0][o], l, mid, L, R, k);

    if (R > mid) add(ch[1][o], mid + 1, r, L, R, k);

    pushup(o, l, r);

}

void merge(int &o, int u, int l, int r) {

    if (!o || !u) {

        if (!o) o = u;

        return ;

    }

    addv[o] += addv[u];

    if (l < r) {

        merge(ch[0][o], ch[0][u], l, mid);

        merge(ch[1][o], ch[1][u], mid + 1, r);

    }

    pushup(o, l, r);

}

void dfs(int u) {

    for (int i = 0, v; i < G[u].size(); ++i) {

        v = G[u][i];

        if (v == fa[u]) continue;

        dfs(v);

        merge(root[u], root[v], 1, n);

    }

    for (int i = 0; i < cf[u].size(); ++i)

        for (int j = 0; j < segs[cf[u][i].first].size(); ++j)

            add(root[u], 1, n, segs[cf[u][i].first][j].first, segs[cf[u][i].first][j].second, cf[u][i].second);

    ans += max(0, sumv[root[u]] - 1); // 注意这里要与0取max

}

int main() {

    scanf("%d%d", &n, &m);

    for (int i = 1, x, y; i < n; ++i) {

        scanf("%d%d", &x, &y);

        G[x].pb(y), G[y].pb(x);

    }

    dfs1(1, 0), dfs2(1, 1);

    for (int i = 1, s, t; i <= m; ++i) {

        scanf("%d%d", &s, &t);

        addModify(i, s, t);

    }

    dfs(1);

    ans /= 2;

    printf("%lld\n", ans);

    return 0;

}