HDU 3974 Assign the task (DFS+线段树)

题意：给定一棵树的公司职员管理图，有两种操作，

第一种是 T x y，把 x 及员工都变成 y，

第二种是 C x 询问 x 当前的数。

析：先把该树用dfs遍历，形成一个序列，然后再用线段树进行维护，很简单的线段树。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
 
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 5e4 + 10;
const int mod = 1000000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}
 
vector<int> G[maxn];
int in[maxn], out[maxn];
int cnt;
int sum[maxn<<2], setv[maxn<<2];
 
void dfs(int u){
  in[u] = ++cnt;
  for(int i = 0; i < G[u].size(); ++i)
    dfs(G[u][i]);
  out[u] = cnt;
}
 
void push_down(int rt){
  if(setv[rt] == -1)  return ;
  int l = rt<<1, r = rt<<1|1;
  sum[l] = sum[r] = setv[rt];
  setv[l] = setv[r] = setv[rt];
  setv[rt] = -1;
}
 
void update(int L, int R, int val, int l, int r, int rt){
  if(L <= l && r <= R){
    sum[rt] = val;
    setv[rt] = val;
    return ;
  }
  push_down(rt);
  int m = l + r >> 1;
  if(L <= m)  update(L, R, val, lson);
  if(R > m)   update(L, R, val, rson);
}
 
int query(int M, int l, int r, int rt){
  if(l == r)  return sum[rt];
  push_down(rt);
  int m = l + r >> 1;
  return M <= m ? query(M, lson) : query(M, rson);
}
 
int main(){
  int T;  cin >> T;
  for(int kase = 1; kase <= T; ++kase){
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i)  G[i].clear();
    memset(in, 0, sizeof in);
    for(int i = 1; i < n; ++i){
      int u, v;
      scanf("%d %d", &u, &v);
      G[v].push_back(u);
      ++in[u];
    }
    cnt = 0;
    for(int i = 1; i <= n; ++i)
      if(!in[i]){ dfs(i);  break; }
    memset(setv, -1, sizeof setv);
    memset(sum, -1, sizeof sum);
 
    scanf("%d", &m);
    char s[5];
    printf("Case #%d:\n", kase);
    while(m--){
      int x, y;
      scanf("%s %d", s, &x);
      if(s[0] == 'C')  printf("%d\n", query(in[x], 1, n, 1));
      else {
        scanf("%d", &y);
        update(in[x], out[x], y, 1, n, 1);
      }
    }
  }
  return 0;
}