网站关键词如何部署,什么是软文营销?,涉密网络运行维护服务外包的单位,河北省招投标信息网xor
思路
题目是要求[l,r][l, r][l,r]的所有集合是否都可以得到xxx#xff0c;那么显然我们可以对这[l,r][l, r][l,r]个线性基求交#xff0c;然后再特判能否xxx能否插入#xff0c;如果能插入#xff0c;显然输出NONONO#xff0c;否则就输出YESYESYES#xff0c;所以…xor
思路
题目是要求[l,r][l, r][l,r]的所有集合是否都可以得到xxx那么显然我们可以对这[l,r][l, r][l,r]个线性基求交然后再特判能否xxx能否插入如果能插入显然输出NONONO否则就输出YESYESYES所以问题转换成了如何求这[l,r][l, r][l,r]个集合的线性基交了。
有个最简单的方法就是用线段树来维护了然后暴力的得到[l,r][l, r][l,r]中的log(n)log(n)log(n)个线性基交然后再判断是否有集合是无法构成xxx的即可。
代码
/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include bits/stdc.h#define mp make_pair
#define pb push_back
#define endl \n
#define mid (l r 1)
#define lson rt 1, l, mid
#define rson rt 1 | 1, mid 1, r
#define ls rt 1
#define rs rt 1 | 1using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pairint, int pii;const double pi acos(-1.0);
const double eps 1e-7;
const int inf 0x3f3f3f3f;inline ll read() {ll f 1, x 0;char c getchar();while(c 0 || c 9) {if(c -) f -1;c getchar();}while(c 0 c 9) {x (x 1) (x 3) (c ^ 48);c getchar();}return f * x;
}// typedef unsigned int ui;const int N 5e4 10;struct LinearBasis {ll base[35];void init() {memset(base, 0, sizeof base);}ll operator [] (int pos) {return base[pos];}bool insert(ll x) {for(int i 31; i 0; i--) {if(x i 1) {if(!base[i]) {base[i] x;return true;}x ^ base[i];}}return false;}bool judge(ll x) {for(int i 31; i 0; i--) {if(x i 1) {if(!base[i]) {return true;}x ^ base[i];}}return false;}LinearBasis inter (const LinearBasis t) {LinearBasis ans, c t, d t;ans.init();for(int i 0; i 32; i) {if(!base[i]) continue;int p i;ll x base[i], temp 0;for(int j p; j 0; j--) {if(x j 1) {if(c[j]) {x ^ c[j]; temp ^ d[j];}else {p j; break;}}}if(!x) {ans[i] temp;}else {c[p] x; d[p] temp;}}return ans;}
}tree[N 2];void push_up(int rt) {tree[rt] tree[ls].inter(tree[rs]);
}void build(int rt, int l, int r) {if(l r) {int n read();for(int i 1; i n; i) {ll x read();tree[rt].insert(x);}return ;}build(lson);build(rson);push_up(rt);
}int flag;void query(int rt, int l, int r, int L, int R, ll x) {if(l L r R) {if(tree[rt].judge(x)) flag 0;return ;}if(L mid) query(lson, L, R, x);if(R mid) query(rson, L, R, x);
}int main() {// freopen(in.txt, r, stdin);// freopen(out.txt, w, stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int n read(), m read();build(1, 1, n);for(int i 1; i m; i) {int l read(), r read(); ll x read(); flag 1;query(1, 1, n, l, r, x);puts(flag ? YES : NO);}return 0;
}
