51nod 1681 公共祖先

先遍历第一棵树，按DFS序给每个节点标号，并记录每棵子树标号所处的范围的左端点 l[i] 和右端点 r[i]。
然后遍历第二棵树，分别记录遍历一棵子树前后 l[i]~r[i] 范围内有多少点被遍历过，设二者之差为x，ans += x * (x - 1) / 2 即可。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
template <class T>
void read(T &x){
char c;
while(c = getchar(), c < '0' || c > '9');
x = c - '0';
while(c = getchar(), c >= '0' && c <= '9')
x = x * 10 + c - '0';
}
template <class T>
void write(T x){
if(x >= 10) write(x / 10);
putchar('0' + x % 10);
}

const int N = 100005;
int n, l
, r
, idx, sum
;
int ecnt, adj[2]
, nxt[2]
, go[2]
;
ll ans;
bool notroot
;

void add(int u, int v, bool x){
notroot[v] = 1;
go[x][++ecnt] = v;
nxt[x][ecnt] = adj[x][u];
adj[x][u] = ecnt;
}
void dfs1(int u, int pre){
l[u] = ++idx;
for(int e = adj[0][u]; e; e = nxt[0][e])
if(go[0][e] != pre)
dfs1(go[0][e], u);
r[u] = idx;
}
void change(int p){
while(p <= n) sum[p]++, p += p & -p;
}
int ask(int p){
int ret = 0;
while(p) ret += sum[p], p -= p & -p;
return ret;
}
void dfs2(int u, int pre){
change(l[u]);
int old = ask(r[u]) - ask(l[u] - 1);
for(int e = adj[1][u]; e; e = nxt[1][e])
if(go[1][e] != pre)
dfs2(go[1][e], u);
int now = ask(r[u]) - ask(l[u] - 1);
int delta = now - old;
ans += (ll) delta * (delta - 1) / 2;
}

int main(){

read(n);
for(int i = 1, u, v; i < n; i++)
read(u), read(v), add(u, v, 0);
for(int i = 1; i <= n; i++)
if(!notroot[i])
dfs1(i, 0);
memset(notroot, 0, sizeof(notroot));
for(int i = 1, u, v; i < n; i++)
read(u), read(v), add(u, v, 1);
for(int i = 1; i <= n; i++)
if(!notroot[i])
dfs2(i, 0);
write(ans), putchar('\n');

return 0;
}