题解 无向图

传送门

发现就是要动态维护一个集合线性基的大小
线段树分治即可

点击查看代码

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 100010
#define fir first
#define sec second
#define pb push_back
#define ll long long
//#define int long long

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int m, q;

namespace force{
	set<int> s;
	int sta[N], uni[N], usiz, top;
	int recalc() {
		top=usiz=0;
		for (auto it:s) sta[top++]=it;
		int lim=1<<top;
		for (int s=0; s<lim; ++s) {
			int sum=0;
			for (int i=0; i<top; ++i) if (s&(1<<i)) sum^=sta[i];
			uni[++usiz]=sum;
		}
		sort(uni+1, uni+usiz+1);
		usiz=unique(uni+1, uni+usiz+1)-uni-1;
		return usiz;
	}
	void solve() {
		for (int i=1,x; i<=q; ++i) {
			if (read()&1) {
				x=read();
				s.insert(x);
				printf("%d\n", recalc());
			}
			else {
				x=read();
				s.erase(x);
				printf("%d\n", recalc());
			}
		}
	}
}

namespace task1{
	ll ans[N];
	map<int, int> mp;
	vector<int> add[N<<2];
	int tl[N<<2], tr[N<<2];
	#define tl(p) tl[p]
	#define tr(p) tr[p]
	#define dat(p) dat[p]
	struct basic{
		int base[32], siz;
		basic() {siz=0; memset(base, 0, sizeof(base));}
		void insert(int t) {
			for (int i=31; ~i; --i) if (t&(1<<i)) {
				if (!base[i]) {base[i]=t; ++siz; return ;}
				else t^=base[i];
			}
		}
	}dat[N<<2];
	void build(int p, int l, int r) {
		tl(p)=l; tr(p)=r;
		if (l==r) return ;
		int mid=(l+r)>>1;
		build(p<<1, l, mid);
		build(p<<1|1, mid+1, r);
	}
	void upd(int p, int l, int r, int val) {
		if (l<=tl(p)&&r>=tr(p)) {add[p].pb(val); return ;}
		int mid=(tl(p)+tr(p))>>1;
		if (l<=mid) upd(p<<1, l, r, val);
		if (r>mid) upd(p<<1|1, l, r, val);
	}
	void query(int p, basic base) {
		for (auto it:add[p]) base.insert(it);
		if (tl(p)==tr(p)) {ans[tl(p)]=1ll<<base.siz; return ;}
		query(p<<1, base); query(p<<1|1, base);
	}
	void solve() {
		build(1, 1, q);
		for (int i=1,x; i<=q; ++i) {
			if (read()&1) mp[read()]=i;
			else {
				x=read();
				upd(1, mp[x], i-1, x);
				mp.erase(x);
			}
		}
		for (auto it:mp) upd(1, it.sec, q, it.fir);
		query(1, basic());
		for (int i=1; i<=q; ++i) printf("%lld\n", ans[i]);
	}
}

signed main()
{
	freopen("xor.in", "r", stdin);
	freopen("xor.out", "w", stdout);
	
	m=read(); q=read();
	// force::solve();
	task1::solve();

	return 0;
}