CF 570 D. Tree Requests

D. Tree Requests

http://codeforces.com/problemset/problem/570/D

题意：

　　一个以1为根的树，每个点上有一个字母(a-z)，每次询问一个子树内深度为h的点是否可以构成回文串。（深度是到1的深度，没有也算，空回文串）

分析：

　　dsu on tree。询问子树信息。

　　判断是否构成回文：出现奇数次的字符小于等于1个。

代码：

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<iostream>
 6 #include<cctype>
 7 #include<set>
 8 #include<vector>
 9 #include<queue>
10 #include<map>
11 #define pa pair<int,int>
12 #define mp(a,b) make_pair(a,b)
13 using namespace std;
14 typedef long long LL;
15 
16 inline int read() {
17     int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
18     for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
19 }
20 
21 const int N = 500005;
22 
23 int head[N], nxt[N], to[N], En;
24 int fa[N], siz[N], son[N], deth[N], ans[N], cnt[N][27], ch[N];
25 char s[N];
26 vector< pa > q[N];
27 
28 void add_edge(int u,int v) {
29     ++En; to[En] = v; nxt[En] = head[u]; head[u] = En;
30 }
31 
32 void dfs(int u,int fa) {
33     siz[u] = 1;
34     deth[u] = deth[fa] + 1;
35     for (int i=head[u]; i; i=nxt[i]) {
36         int v = to[i];
37         dfs(v, u);
38         siz[u] += siz[v];
39         if (!son[u] || siz[son[u]] < siz[v]) son[u] = v;
40     }
41 }
42 
43 void add(int u) {
44     cnt[deth[u]][ch[u]] ++;
45 }
46 void Calc(int u) {
47     add(u);
48     for (int i=head[u]; i; i=nxt[i]) Calc(to[i]);
49 }
50 void Clear(int u) {
51     cnt[deth[u]][ch[u]] --;
52     for (int i=head[u]; i; i=nxt[i]) Clear(to[i]);
53 }
54 
55 void solve(int u,bool c) {
56     for (int i=head[u]; i; i=nxt[i]) 
57         if (to[i] != son[u]) solve(to[i], 0);
58     if (son[u]) solve(son[u], 1);
59     
60     for (int i=head[u]; i; i=nxt[i]) 
61         if (to[i] != son[u]) Calc(to[i]);
62     add(u);
63     
64     for (int i=0,sz=q[u].size(); i<sz; ++i) {
65         int flag = 0, id = q[u][i].second, h = q[u][i].first;
66         for (int j=0; j<26; ++j) if (cnt[h][j] & 1) flag ++;
67         ans[id] = (flag <= 1);
68     }
69     
70     if (!c) Clear(u);
71 }
72 
73 int main() {
74     int n = read(), Q = read();
75     for (int i=2; i<=n; ++i) {
76         int u = read();
77         add_edge(u, i);
78     }
79     scanf("%s",s + 1);
80     for (int i=1; i<=n; ++i) ch[i] = s[i] - 'a';
81     for (int i=1; i<=Q; ++i) {
82         int v = read(), h = read();
83         q[v].push_back(mp(h, i));
84     }
85     dfs(1, 0);
86     solve(1, 1);
87     for (int i=1; i<=Q; ++i) puts(ans[i] ? "Yes" : "No");
88     return 0;
89 }