[题解]CF1899G Unusual Entertainment

思路

比较典的一个题。

把一棵树的 DFS 序剖下来过后再同一棵子树内的节点编号是连续的。

那么，我们查询 \((l,r,x)\) 时，问题就可以转化为在 \(p_{l \sim r}\) 中是否出现过 \(lid_x \sim rid_x\) 中的数，其中 \(lid_i\) 表示 \(i\) 所在子树的节点最小的 DFS 序的编号，\(rid_i\) 同理。

然后把询问离线下来做即可。

Code

#include <bits/stdc++.h>  
#define re register  
  
using namespace std;  
  
const int N = 2e5 + 10,M = 4e5 + 10;  
int n,q;  
int ans[2][N];  
int idx,h[N],ne[M],e[M],arr[N];  
int num,id[N],sz[N];  
  
struct Query{  
    int op,l,r,id;  
};  
vector<Query> Q[N];  
  
inline int read(){  
    int r = 0,w = 1;  
    char c = getchar();  
    while (c < '0' || c > '9'){  
        if (c == '-') w = -1;  
        c = getchar();  
    }  
    while (c >= '0' && c <= '9'){  
        r = (r << 3) + (r << 1) + (c ^ 48);  
        c = getchar();  
    }  
    return r * w;  
}  
  
inline void add(int a,int b){  
    ne[idx] = h[a];  
    e[idx] = b;  
    h[a] = idx++;  
}  
  
struct BIT{  
    int tr[N];  
  
    inline int lowbit(int x){  
        return x & -x;  
    }  
  
    inline void modify(int x,int k){  
        for (re int i = x;i <= n;i += lowbit(i)) tr[i] += k;  
    }  
  
    inline int query(int x){  
        int res = 0;  
        for (re int i = x;i;i -= lowbit(i)) res += tr[i];  
        return res;  
    }  
}T;  
  
inline void init(){  
    for (re int i = 1;i <= n;i++){  
        Q[i].clear();  
        h[i] = -1;  
        T.tr[i] = ans[0][i] = ans[1][i] = sz[i] = id[i] = 0;  
    }  
}  
  
inline void dfs(int u,int fa){  
    sz[u] = 1;  
    id[u] = ++num;  
    for (re int i = h[u];~i;i = ne[i]){  
        int j = e[i];  
        if (j == fa) continue;  
        dfs(j,u);  
        sz[u] += sz[j];  
   }  
}  
  
inline void solve(){  
    idx = num = 0;  
    n = read();  
    q = read();  
    init();  
    for (re int i = 1;i < n;i++){  
        int a,b;  
        a = read();  
        b = read();  
        add(a,b);  
        add(b,a);  
    }  
    dfs(1,0);  
    for (re int i = 1;i <= n;i++) arr[i] = read();  
    for (re int i = 1;i <= q;i++){  
        int l,r,x;  
        l = read();  
        r = read();  
        x = read();  
        Q[l - 1].push_back({0,id[x],id[x] + sz[x] - 1,i});  
        Q[r].push_back({1,id[x],id[x] + sz[x] - 1,i});  
    }  
    for (re int i = 1;i <= n;i++){  
        T.modify(id[arr[i]],1);  
        for (auto p:Q[i]) ans[p.op][p.id] = T.query(p.r) - T.query(p.l - 1);  
    }  
    for (re int i = 1;i <= q;i++){  
        if (ans[1][i] - ans[0][i]) puts("YES");  
        else puts("NO");  
    }  
}  
  
int main(){  
    int T;  
    T = read();  
    while (T--) solve();  
    return 0;  
}