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{sumlimits_{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
", ans);
return 0;
}