题目链接

题意：对于m次询问 求解以vi为根节点 深度为hi的的字母能不能组合成回文串。

思路：暴力dsu找一边 简直就是神技！

#include
#define ll long long int
using namespace std;
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
int moth[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]={1,0 ,0,1 ,-1,0 ,0,-1};
int dirs[8][2]={1,0 ,0,1 ,-1,0 ,0,-1, -1,-1 ,-1,1 ,1,-1 ,1,1};
const int inf=0x3f3f3f3f;
const int N=500007;
const ll mod=1e9+7;
struct node{
int h,id;
bool f;
friend bool operator < (node a,node b){ return a.id G[N];
char val[N];
int dp[N],son[N],cnt[N][26];
int odd[N];
int de[N];
int po=0;
vector h[N];
node ans[N];
void dfs(int u,int fa,int deep){
dp[u]=1;
de[u]=deep;
for(int i=0;idp[son[u]]) son[u]=v; //树剖
}
}
void add(int u,int fa,int w,int deep){
cnt[deep][val[u]-'a']+=w;
if(cnt[deep][val[u]-'a']&1) odd[deep]++;
else odd[deep]--;
for(int i=0;i=2) tmp.f=0;
ans[++num]=tmp;
}
if(!opt) add(u,fa,-1,deep);
}
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
int n,m; cin>>n>>m;
for(int i=2;i<=n;i++){ int fa; cin>>fa;
G[fa].push_back(i);
}
for(int i=1;i<=n;i++){ cin>>val[i];
}
for(int i=1;i<=m;i++){ int vi,hi; cin>>vi>>hi;
node tmp; tmp.h=hi; tmp.f=1; tmp.id=i;
h[vi].push_back(tmp);
}
dfs(1,0,1);
dfss(1,0,1,1);
sort(ans+1,ans+1+m);
for(int i=1;i<=num;i++){
if(ans[i].f) cout<<"Yes\n";
else cout<<"No\n";
}
return 0;
}