2<= n <=300000
题意：就是按顺序走这些路，求每个点被经过几次。
思路：点的差分，没啥好说的，就是从第二个点之后经过次数都要减一，因为后面的点又做起点又做终点，就会被计多一次， 而最后一个点减一就是由于题意说的。
#include <bits/stdc++.h> using namespace std; inline int read() { int x = 0, f = 1; char ch = getchar(); while (ch < ‘0‘ || ch > ‘9‘) { if (ch == ‘-‘) f = -1; ch = getchar(); } while (ch >= ‘0‘ && ch <= ‘9‘) { x = x * 10 + ch - 48; ch = getchar(); } return x * f; } const int N = 300000 + 100; struct E { int v, ne; } e[N << 1]; int head[N], cnt, s[N], lca[N][25], a[N], dep[N], n, fa[N]; bool vis[N]; inline void add(int u, int v) { e[++cnt].v = v; e[cnt].ne = head[u]; head[u] = cnt; } void dfs(int u) { vis[u] = 1; lca[u][0] = fa[u]; for (int i = 1; i <= 24; i++) lca[u][i] = lca[lca[u][i-1]][i-1]; for (int i = head[u]; i; i = e[i].ne) { int v = e[i].v; if (vis[v]) continue; dep[v] = dep[u] + 1; fa[v] = u; dfs(v); } } void getans(int u) { for (int i = head[u]; i; i = e[i].ne) { int v = e[i].v; if (v == lca[u][0]) continue; getans(v); s[u] += s[v]; } } int Lca(int u, int v) { if (dep[u] < dep[v]) swap(u, v); for (int i = 24; i >= 0; i--) if (dep[lca[u][i]] >= dep[v]) u = lca[u][i]; if (u == v) return u; for (int i = 24; i >= 0; i--) if (lca[u][i] != lca[v][i]) u = lca[u][i], v = lca[v][i]; return lca[u][0]; } int main() { n = read(); for (int i = 1; i <= n; i++) a[i] = read(); for (int i = 1; i < n; i++) { int u = read(), v = read(); add(u, v); add(v, u); } dep[1] = 1; dfs(1); for (int i = 1; i <= n - 1; i++) { int u = a[i], v = a[i + 1]; int p = Lca(u, v); //printf("%d %d %d %d\n", u, v, p, lca[p][0]); s[u]++, s[v]++; s[p]--, s[lca[p][0]]--; } getans(1); for (int i = 2; i <= n; i++) s[a[i]]--; for (int i = 1; i <= n; i++) printf("%d\n", s[i]); return 0; }
原文：https://www.cnblogs.com/Mrzdtz220/p/11027858.html
