The first line contains a single positive integer T( T <= 10 ), indicates the number of test cases.
For each test case:
The first line contains an integer N (N ≤ 50,000) , which is the number of the employees.
The following N - 1 lines each contain two integers u and v, which means the employee v is the immediate boss of employee u(1<=u,v<=N).
The next line contains an integer M (M ≤ 50,000).
The following M lines each contain a message which is either
"C x" which means an inquiry for the current task of employee x
or
"T x y"which means the company assign task y to employee x.
(1<=x<=N,0<=y<=10^9)
For each test case, print the test case number (beginning with 1) in the first line and then for every inquiry, output the correspond answer per line.
首先，读者可以画个树，自行标号dfs序，体会下dfs序的特点
不难发现，对于任意子树所含的结点编号是连续的
且该子树所含结点最小编号就是根结点编号
然后这个dfs序就是我们解决这道题的关键所在
按题意直接建好多叉树，标好dfs序，然后你会发现其实操作蛮简单的
查询员工，变成一层层去找对应子树根结点
#include <cstdio> #include <set> #include <cstring> using namespace std; const int maxn = 50005; struct node{int dat, child, nxt, tag, boss, l, r;}tr[maxn]; int t, n, m, boss, tot, p[maxn], a, b; void build(int p) { tr[p].l = ++tot; for (int i = tr[p].child; i; i = tr[i].nxt)build(i); tr[p].r = tot; } inline void create(int Boss, int k) { tr[k].boss = Boss; tr[k].nxt = tr[Boss].child; tr[Boss].child = k; } inline void push(int p) { if (!tr[p].tag)return; for (int i = tr[p].child; i; i = tr[i].nxt) tr[i].dat = tr[i].tag = tr[p].tag; tr[p].tag = 0; } inline void update(int l, int r, int p) { if (l == p) { tr[p].dat = tr[p].tag = r; return; } push(p); for (int i = tr[p].child; i;i=tr[i].nxt) if (tr[i].l <= tr[l].l && tr[l].l <= tr[i].r) update(l, r, i); } inline int query(int x, int p) { if (x == p) return tr[p].dat?tr[p].dat:-1; push(p); for (int i = tr[p].child; i; i = tr[i].nxt) if (tr[i].l <= tr[x].l && tr[x].l <= tr[i].r) return query(x, i); } int main() { scanf("%d", &t); for (int k = 1; k <= t; ++k) { scanf("%d", &n); printf("Case #%d:\n", k); memset(tr, 0, sizeof tr); for (int i = 1; i < n; ++i) { int a, b; scanf("%d%d", &a, &b); create(b, a); } for (int i = 1; i <= n; ++i)if (!tr[i].boss) { boss = i; break; } build(boss); scanf("%d", &m); for (int i = 1; i <= m; ++i) { char s[2]; scanf("%s%d", s, &a); if (s[0] == ‘C‘)printf("%d\n", query(a, boss)); else { scanf("%d", &b); update(a, b, boss); } } } }
每次查询，直接往上早，如果boss任务更新时间比当前任务晚，则更新当前任务
#include <cstdio> using namespace std; struct node {int job, time;}a[50005]; int fa[50005], t, n, m, u, v, cnt; char c[2]; int main() { scanf("%d", &t); for (int cas = 1, cnt = 0; cas <= t; ++cas, cnt = 0) { scanf("%d",&n); for (int i = 1; i <= n; ++i) fa[i] = a[i].job = a[i].time = -1; for (int i = 1; i < n; ++i) scanf("%d%d",&u,&v), fa[u] = v; scanf("%d", &m); printf("Case #%d:", cas); while (m--) { int x; scanf("%s%d", c, x); if (c[0] == ‘C‘) { int y = a[x].job, tmp = a[x].time; for (;x != -1;x=fa[x]) if (a[x].time > tmp) y = a[x].job, tmp = a[x].time; printf("%d\n", &y); } else { int y; scanf("%d", &y); a[x].job = y, a[x].time = ++cnt; } } } return 0; }
原文：https://www.cnblogs.com/2aptx4869/p/12123748.html