2019年牛客多校第四场 B题xor(线段树+线性基交)

题目链接

传送门

题意

给你\(n\)个基底，求\([l,r]\)内的每个基底是否都能异或出\(x\)。

思路

线性基交板子题，但是一直没看懂咋求，先偷一份咖啡鸡板子写篇博客吧~

线性基交学习博客：传送门

代码实现如下

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;

typedef unsigned int ui;
typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;

#define lson rt<<1
#define rson rt<<1|1
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("D://Code//in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define  fuck(x)     cout<<#x" = "<<x<<endl

const double eps = 1e-8;
const int mod = 1000000007;
const int maxn = 50000 + 7;
const double pi = acos ( -1 );
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;

ui x, a[maxn][35];
int n, q, sz, l, r;

struct base{
    ui r[32];
    ui f[32];
    bool ins(ui x){
        for (int i=31;i>=0;i--)
            if (x>>i){
                if (!r[i]) {r[i]=x;return 1;}
                x^=r[i];
                if (!x) return 0;
            }
        return 0;
    }
    void ins2(ui x){
        ui tmp=x;
        for (int i=31;i>=0;i--)
            if (x>>i){
                if (!r[i]) {f[i]=tmp;r[i]=x;return;}
                x^=r[i]; tmp^=f[i];
                if (!x) return;
            }
        return;
    }
    bool find(ui x){
        for (int i=31;i>=0;i--)
            if (x>>i){
                if (!r[i]) return 0;
                x^=r[i];
            }
        return x==0;
    }
    ui calc(ui x){
        ui ret=0;
        for (int i=31;i>=0;i--){
            if (x>>i){
                ret^=f[i];
                x^=r[i];
            }
        }
        return ret;
    }
    void print(){
        for (int i=0;i<32;i++)cout<<r[i]<<' ';cout<<endl;
    }
    void clear(){
        for (int i=0;i<32;i++) r[i]=f[i]=0;
    }
};

struct node {
    int l, r;
    base val;
}segtree[maxn<<2];

void push_up(int rt) {
    base tmp = segtree[lson].val;
    base ans;
    ans.clear();
    for(int i = 31; i >= 0; --i) {
        ui x = segtree[rson].val.r[i];
        if(tmp.find(x)) {
            ans.ins(x^tmp.calc(x));
        } else tmp.ins2(x);
    }
    segtree[rt].val = ans;
}

void build(int rt, int l, int r) {
    segtree[rt].l = l, segtree[rt].r = r;
    if(l == r) {
        for(int i = 0; i <= 31; ++i) segtree[rt].val.ins(a[l][i]);
        return;
    }
    int mid = (l + r) >> 1;
    build(lson, l, mid);
    build(rson, mid + 1, r);
    push_up(rt);
}

bool query(int rt, int l, int r, LL x) {
    if(segtree[rt].l == l && segtree[rt].r == r) {
        return segtree[rt].val.find(x);
    }
    int mid = (segtree[rt].l + segtree[rt].r) >> 1;
    if(r <= mid) return query(lson, l, r, x);
    else if(l > mid) return query(rson, l, r, x);
    else {
        return query(lson, l, mid, x) && query(rson, mid + 1, r, x);
    }
}

int main() {
#ifndef ONLINE_JUDGE
FIN;
#endif // ONLINE_JUDGE
    scanf("%d%d", &n, &q);
    for(int i = 1; i <= n; ++i) {
        scanf("%d", &sz);
        for(int j = 0; j < sz; ++j) scanf("%lld", &a[i][j]);
        for(int j = sz; j <= 31; ++j) a[i][j] = 0;
    }
    build(1, 1, n);
    while(q--) {
        scanf("%d%d%lld", &l, &r, &x);
        if(query(1, l, r, x)) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}