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n個點的樹，以點1為根，告訴你經過每一秒，某頂點上的蘋果會到達哪個頂點，且已知一個頂點每同時出現兩個蘋果就會碰撞消失（如2得0，如3得1），求有多少個蘋果能到達頂點1

bfs求點深度

 1 #include <queue>
 2 #include <vector>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <algorithm>
 6 using namespace std;
 7 typedef long long LL;
 8 typedef pair<int
 
, int> P;
 9 const int maxn = 1e5 + 10;
10 bool ans[maxn];
11 bool vis[maxn];
12 int n;
13 
14 vector<int> G[maxn];
15 
16 int main() {
17     scanf("%d", &n);
18     for (int i = 2; i <= n; i++) {
19         int u;
20         scanf("%d", &u);
21         G[u].push_back(i);
22     }
 
23     queue<P> que;
24     que.push(P(1, 0));
25     ans[0] = 1;
26     while (!que.empty()) {
27         P p = que.front(); que.pop();
28         int v = p.first;
29         int dis = p.second;
30         for (int i = 0; i < G[v].size(); i++) {
31             int tmp = G[v][i];
32             que.push(P(tmp, dis + 1
 
));
33             ans[dis + 1] ^= 1;
34         }
35     }
36     int res = 0;
37     for (int i = 0; i <= n; i++) {
38         res += ans[i];
39     }
40     printf("%d\n", res);
41     return 0;
42 }

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