HDU 3974 Assign the task

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dfs序可以把结点v及其子树放在连续的区间中
分配任务操作T x y相当于把一个连续的区间置为y，可以用线段树来维护
查询操作为线段树的单点查询

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#include <bits/stdc++.h>

using namespace std;

const int MAX_N = 50000 + 10;

int N, M;
vector<int> G[MAX_N];
bool leader[MAX_N];

vector<int> vs;
int id[MAX_N];
int size[MAX_N];

void init() {
  for (int v = 1; v <= N; v++) {
    G[v].clear();
    leader[v] = true;
  }
  vs.clear();
}

void dfs(int v) {
  id[v] = vs.size();
  vs.push_back(v);
  size[v] = 1;
  for (int i = 0; i < G[v].size(); i++) {
    int u = G[v][i];
    dfs(u);
    size[v] += size[u];
  }
}

#define lc (2 * k + 1)
#define rc (2 * k + 2)
#define m ((l + r) / 2)

const int ST_SIZE = 1 << 17;
int task[ST_SIZE];

void build() {
  memset(task, -1, sizeof(task));
}

void push_down(int k) {
  if (task[k] != -1) {
    task[lc] = task[rc] = task[k];
    task[k] = -1;
  }
}

void update(int k, int l, int r, int a, int b, int x) {
  if (b <= l || r <= a) {
    return;
  } else if (a <= l && r <= b) {
    task[k] = x;
  } else {
    push_down(k);
    update(lc, l, m, a, b, x);
    update(rc, m, r, a, b, x);
  }
}

int query(int k, int l, int r, int a) {
  if (l == r - 1) {
    return task[k];
  } else {
    push_down(k);
    if (a < m) return query(lc, l, m, a);
    else return query(rc, m, r, a);
  }
}

int main() {
  int T;
  scanf("%d", &T);
  for (int t = 1; t <= T; t++) {
    printf("Case #%d:\n", t);
    scanf("%d", &N);
    init();
    for (int i = 1; i < N; i++) {
      int u, v;
      scanf("%d %d", &u, &v);
      G[v].push_back(u);
      leader[u] = false;
    }

    int root = -1;
    for (int v = 1; v <= N; v++) {
      if (leader[v]) root = v;
    }

    dfs(root);
    build();

    scanf("%d", &M);
    for (int i = 0; i < M; i++) {
      char cmd;
      int x, y;
      scanf(" %c", &cmd);
      if (cmd == 'C') {
        scanf("%d", &x);
        printf("%d\n", query(0, 0, N, id[x]));
      } else {
        scanf("%d %d", &x, &y);
        update(0, 0, N, id[x], id[x] + size[x], y);
      }

    }
  }
  return 0;
}