\(\Large\textbf{Description:} \large{有一颗 n 个节点的树，k 次旅行，问每一条边被走过的次数。}\)
\(\Large\textbf{Solution:} \large{树上差分板子？？}\)
\(\Large\textbf{Code:}\)
#include
#include
#define LL long long
#define gc() getchar()
#define rep(i, a, b) for (int i = (a); i <= (b); ++i)
using namespace std;
const int N = 1e5 + 5;
int n, k, cnt, head[N], u[N], v[N], ans[N];
int size[N], top[N], dep[N], fa[N], son[N], w[N];
struct Edge {
int to, next;
}e[N << 1];
inline int read() {
char ch = gc();
int ans = 0, flag = 1;
while (ch > '9' || ch < '0') {
if (ch == '-') flag = -1;
ch = gc();
}
while (ch >= '0' && ch <= '9') ans = (ans << 1) + (ans << 3) + ch - '0', ch = gc();
return ans;
}
inline void add(int x, int y) {
e[++cnt].to = y;
e[cnt].next = head[x];
head[x] = cnt;
}
inline void dfs1(int x, int y) {
int Max = -1;
dep[x] = dep[y] + 1;
fa[x] = y;
size[x] = 1;
for (int i = head[x]; i ; i = e[i].next) {
int U = e[i].to;
if (U == y) continue;
dfs1(U, x);
size[x] += size[U];
if (size[U] > Max) Max = size[U], son[x] = U;
}
}
inline void dfs2(int x, int y) {
top[x] = y;
if (!son[x]) return ;
dfs2(son[x], y);
for (int i = head[x]; i ; i = e[i].next) {
int U = e[i].to;
if (U == fa[x] || U == son[x]) continue;
dfs2(U, U);
}
}
inline int lca(int x, int y) {
while (top[x] != top[y]) {
if (dep[top[x]] < dep[top[y]]) swap(x, y);
x = fa[top[x]];
}
if (dep[x] > dep[y]) swap(x, y);
return x;
}
inline int dfs(int x) {
int cur = w[x];
for (int i = head[x]; i ; i = e[i].next) {
int U = e[i].to;
if (U == fa[x]) continue;
cur += dfs(U);
}
return ans[x] = cur;
}
int main() {
n = read();
rep(i, 1, n - 1) { u[i] = read(); v[i] = read(); add(u[i], v[i]); add(v[i], u[i]); }
dfs1(1, 0);//树剖 方便求lca。
dfs2(1, 1);
k = read();
int l, r;
rep(i, 1, k) l = read(), r = read(), ++w[l], ++w[r], w[lca(l, r)] -= 2;//边的差分。
dfs(1);//最后遍历一遍树，每条边的走过次数就是它的子树和。
rep(i, 1, n - 1) { if (dep[u[i]] > dep[v[i]]) swap(u[i], v[i]); printf("%d ", ans[v[i]]); }//差分时我们以每条边深度更深的那个点存储这条边的值。
return 0;
} \(\large\color{pink}{by\quad Miraclys}\)