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：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<iostream>
#include<cctype>
#include<set>
#include<vector>
#include<queue>
#include<map>
#define pa pair<int,int>
#define mp(a,b) make_pair(a,b)
using namespace std;
typedef long long LL;

inline int read() {
    int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}

const int N = 500005;

int head[N], nxt[N], to[N], En;
int fa[N], siz[N], son[N], deth[N], ans[N], cnt[N][27], ch[N];
char s[N];
vector< pa > q[N];

void add_edge(int u,int v) {
    ++En; to[En] = v; nxt[En] = head[u]; head[u] = En;
}

void dfs(int u,int fa) {
    siz[u] = 1;
    deth[u] = deth[fa] + 1;
    for (int i=head[u]; i; i=nxt[i]) {
        int v = to[i];
        dfs(v, u);
        siz[u] += siz[v];
        if (!son[u] || siz[son[u]] < siz[v]) son[u] = v;
    }
}

void add(int u) {
    cnt[deth[u]][ch[u]] ++;
}
void Calc(int u) {
    add(u);
    for (int i=head[u]; i; i=nxt[i]) Calc(to[i]);
}
void Clear(int u) {
    cnt[deth[u]][ch[u]] --;
    for (int i=head[u]; i; i=nxt[i]) Clear(to[i]);
}

void solve(int u,bool c) {
    for (int i=head[u]; i; i=nxt[i]) 
        if (to[i] != son[u]) solve(to[i], 0);
    if (son[u]) solve(son[u], 1);
    
    for (int i=head[u]; i; i=nxt[i]) 
        if (to[i] != son[u]) Calc(to[i]);
    add(u);
    
    for (int i=0,sz=q[u].size(); i<sz; ++i) {
        int flag = 0, id = q[u][i].second, h = q[u][i].first;
        for (int j=0; j<26; ++j) if (cnt[h][j] & 1) flag ++;
        ans[id] = (flag <= 1);
    }
    
    if (!c) Clear(u);
}

int main() {
    int n = read(), Q = read();
    for (int i=2; i<=n; ++i) {
        int u = read();
        add_edge(u, i);
    }
    scanf("%s",s + 1);
    for (int i=1; i<=n; ++i) ch[i] = s[i] - 'a';
    for (int i=1; i<=Q; ++i) {
        int v = read(), h = read();
        q[v].push_back(mp(h, i));
    }
    dfs(1, 0);
    solve(1, 1);
    for (int i=1; i<=Q; ++i) puts(ans[i] ? "Yes" : "No");
    return 0;
}