[CF191C]Fools and Roads

题目大意：有一颗$n$个节点的树，$k$次旅行，问每一条被走过的次数。

题解：树上差分，$num_x$表示连接$x$和$fa_x$的边被走过的次数，一条路径$u->v$，$num_u+1,num_v+1,num_{LCA(u,v)}-2$，最后求个子树和就行了

卡点：无

C++ Code：

#include <cstdio>
#include <algorithm>
#define maxn 100010

int head[maxn], cnt = 1;
struct Edge {
	int to, nxt;
} e[maxn << 1];
inline void add(int a, int b) {
	e[++cnt] = (Edge) {b, head[a]}; head[a] = cnt;
}

#define M 18
int dep[maxn], fa[M][maxn];
void dfs(int u) {
	for (int i = 1; i < M; i++) fa[i][u] = fa[i - 1][fa[i - 1][u]];
	for (int i = head[u]; i; i = e[i].nxt) {
		int v = e[i].to;
		if (!dep[v]) {
			dep[v] = dep[u] + 1;
			fa[0][v] = u;
			dfs(v);
		}
	}
}
inline int LCA(int x, int y) {
	if (x == y) return x;
	if (dep[x] < dep[y]) std::swap(x, y);
	for (int i = dep[x] - dep[y]; i; i &= i - 1) x = fa[__builtin_ctz(i)][x];
	if (x == y) return x;
	for (int i = M - 1; ~i; i--) if (fa[i][x] != fa[i][y]) x = fa[i][x], y = fa[i][y];
	return fa[0][x];
}

int V[maxn];
void dfs1(int u) {
	for (int i = head[u]; i; i = e[i].nxt) {
		int v = e[i].to;
		if (v != fa[0][u]) {
			dfs1(v);
			V[u] += V[v];
		}
	}
}

int n, k;
int main() {
	scanf("%d", &n);
	for (int i = 1, a, b; i < n; i++) {
		scanf("%d%d", &a, &b);
		add(a, b);
		add(b, a);
	}
	dep[1] = 1; dfs(1);
	scanf("%d", &k);
	for (int i = 0, a, b; i < k; i++) {
		scanf("%d%d", &a, &b);
		V[a]++, V[b]++, V[LCA(a, b)] -= 2;
	}
	dfs1(1);
	for (int i = 2; i <= cnt; i += 2) {
		int u = e[i ^ 1].to, v = e[i].to;
		if (dep[u] < dep[v]) std::swap(u, v);
		printf("%d", V[u]); putchar((i ^ 1) == cnt ? '\n' : ' ');
	}
	return 0;
}