Bipartite Checking CodeForces - 813F (线段树按时间分治)

大意: 动态添边, 询问是否是二分图.

算是个线段树按时间分治入门题, 并查集维护每个点到根的奇偶性即可.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head

const int N = 1e6+50;
int n, q, fa[N], sz[N], dis[N];
map<pii,vector<int> > g;

vector<pii> tr[N<<2];
void add(int o, int l, int r, int ql, int qr, pii v) {
	if (ql<=l&&r<=qr) return tr[o].pb(v);
	if (mid>=ql) add(ls,ql,qr,v);
	if (mid<qr) add(rs,ql,qr,v);
}

int Find(int x) {while (fa[x]!=x) x=fa[x];return x;}
vector<pair<int*,int> > tag[30];
void build(int o, int l, int r, int d) {
	tag[d].clear();
	for (auto &t:tr[o]) {
		int z = 1, x = t.x, y = t.y;
		while (fa[x]!=x) z^=dis[x],x=fa[x];
		while (fa[y]!=y) z^=dis[y],y=fa[y];
		if (x==y) {
			if (z&1) {
				REP(i,l,r) puts("NO");
				for (auto &t:tag[d]) *t.x=t.y;
				return;
			}
		}
		if (sz[x]<sz[y]) swap(x,y);
		tag[d].pb({&sz[x],sz[x]});
		tag[d].pb({&fa[y],fa[y]});
		tag[d].pb({&dis[y],dis[y]});
		sz[x] += sz[y];
		fa[y] = x, dis[y] = z;
	}
	if (l==r) puts("YES");
	else build(ls,d+1),build(rs,d+1);
	for (auto &t:tag[d]) *t.x = t.y;
}

int main() {
	scanf("%d%d", &n, &q);
	REP(i,1,q) {
		int u ,v;
		scanf("%d%d", &u, &v);
		g[pii(u,v)].pb(i);
	}
	for (auto &t:g) {
		for (int i=0; i<t.y.size(); ++i) {
			if (i+1==t.y.size()) add(1,1,q,t.y[i],q,t.x);
			else {
				add(1,1,q,t.y[i],t.y[i+1]-1,t.x);
				++i;
			}
		}
	}
	REP(i,1,n) fa[i] = i, sz[i] = 1;
	build(1,1,q,0);
}