题意：

传送门

给\(n\)个集合，每个集合有一些数。给出\(m\)个询问，再给出\(l\)和\(r\)和一个数\(v\)，问你任意的\(i \in[l,r]\)的集合，能不能找出子集异或为\(v\)。简单点说，\(v\)能用\([l,r]\)任意一个集合的子集异或和表示。

思路：

子集异或和显然是用线性基。我们用线段树维护任意区间的线性基交集即可。

代码：

/**
求交集 O(logn * logn)
**/
LBasis intersection(const LBasis &a, const LBasis &b){LBasis ans, c = b, d = b;ans.init();for (int i = 0; i <= 32; i++){ll x = a.d[i];if(!x)continue;int j = i;ll T = 0;for(; j >= 0; --j){if((x >> j) & 1)if(c.d[j]) {x ^= c.d[j]; T ^= d.d[j];}else break;}if(!x) ans.d[i] = T;else {c.d[j] = x; d.d[j] = T;}}return ans;
}

#include<map>
#include<set>
#include<cmath>
#include<cstdio>
#include<stack>
#include<ctime>
#include<vector>
#include<queue>
#include<cstring>
#include<string>
#include<sstream>
#include<iostream>
#include<algorithm>
typedef long long ll;
using namespace std;
const int maxn = 50000 + 5;
const int INF = 0x3f3f3f3f;
const ll MOD = 1e9 + 7;
using namespace std;struct LBasis{ll d[33];int tot;void init(){memset(d, 0, sizeof(d));tot = 0;}bool insert(ll x){for(int i = 32; i >= 0; i--){if(x & (1LL << i)){if(d[i]) x ^= d[i];else{d[i] = x;return true;}}}return false;}bool checkin(ll x){for(int i = 32; i >= 0; i--){if(x & (1LL << i)){if(d[i]) x ^= d[i];else return false;}}return true;}};
LBasis intersection(const LBasis &a, const LBasis &b){LBasis ans, c = b, d = b;ans.init();for (int i = 0; i <= 32; i++){ll x = a.d[i];if(!x)continue;int j = i;ll T = 0;for(; j >= 0; --j){if((x >> j) & 1)if(c.d[j]) {x ^= c.d[j]; T ^= d.d[j];}else break;}if(!x) ans.d[i] = T;else {c.d[j] = x; d.d[j] = T;}}return ans;
}
LBasis node[maxn << 2];
void pushup(int rt){node[rt] = intersection(node[rt << 1], node[rt << 1 | 1]);
}
void build(int l, int r, int rt){if(l == r){int sz;scanf("%d", &sz);node[rt].init();while(sz--){ll x;scanf("%lld", &x);node[rt].insert(x);}return;}int m = (l + r) >> 1;build(l, m, rt << 1);build(m + 1, r, rt << 1 | 1);pushup(rt);
}
bool query(int L, int R, int l, int r, ll v, int rt){if(L <= l && R >= r){return node[rt].checkin(v);}int m = (l + r) >> 1;bool ok = true;if(L <= m)ok = ok && query(L, R, l, m, v, rt << 1);if(R > m)ok = ok && query(L, R, m + 1, r, v, rt << 1 | 1);return ok;
}
int main(){int n, m;scanf("%d%d", &n, &m);build(1, n, 1);while(m--){int l, r;ll x;scanf("%d%d%lld", &l, &r, &x);if(query(l, r, 1, n, x, 1)) printf("YES\n");else printf("NO\n");}return 0;
}