bzoj4025二分图(线段树分治 并查集)

/*
思维难度几乎没有， 就是线段树分治check二分图
判断是否为二分图可以通过维护lct看看是否链接出奇环
然后发现不用lct， 并查集维护奇偶性即可

但是复杂度明明一样哈

*/
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#define f1 first
#define f2 second
#define ll long long
#define mmp make_pair
#define lson l, mid, now << 1
#define rson mid + 1, r, now << 1 | 1
#define ls now << 1
#define rs now << 1 | 1
#define M 100010
using namespace std;
int read() {
	int nm = 0, f = 1;
	char c = getchar();
	for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
	return nm * f;
}
vector<pair<int, int> > seg[M << 2];

int n, m, t, father[M], sz[M], ans[M], dis[M];
int find(int x) {
	return father[x] == x ? x : find(father[x]);
}

int getdis(int x) {
	return father[x] == x ? 0 : getdis(father[x]) ^ dis[x];
}
void modify(int l, int r, int now, int ln, int rn, pair<int, int> v) {
	if(l > rn || r < ln) return;
	if(l >= ln && r <= rn) {
		seg[now].push_back(v);
		return;
	}
	int mid = (l + r) >> 1;
	modify(lson, ln, rn, v);
	modify(rson, ln, rn, v);
}

void work(int l, int r, int now) {
	int mid = (l + r) >> 1, f = 0;
	vector<pair<int, int> > tmp;
	for(int i = 0; i < seg[now].size(); i++) {
		int vi = seg[now][i].f1, vj = seg[now][i].f2;
		int a = find(vi), b = find(vj);
		if(a == b) {
			if((getdis(vi) ^ getdis(vj)) == 0) {
				f = 1;
				break;
			}
		} else {
			if(sz[a] > sz[b]) swap(a, b), swap(vi, vj);
			sz[b] += sz[a];
			dis[a] = dis[vi] ^ dis[vj] ^ 1;
			father[a] = b;
			tmp.push_back(mmp(a, b));
		}
	}
	if(!f) {
		if(l == r) ans[l] = 1;
		else work(lson), work(rson);
	}
	for(int i = tmp.size() - 1; i >= 0; i--) {
		int a = tmp[i].first, b = tmp[i].second;
		father[a] = a;
		dis[a] = 0;
		sz[b] -= sz[a];
	}
}
int main() {
	n = read(), m = read(), t = read();
	for(int i = 1; i <= n; i++) father[i] = i, sz[i] = 1;
	for(int i = 1; i <= m; i++) {
		int vi = read(), vj = read(), be = read(), ed = read();
		modify(1, t, 1, be + 1, ed, mmp(vi, vj));
	}
	work(1, t, 1);
	for(int i = 1; i <= t; i++) puts(ans[i] ? "Yes" : "No");
	return 0;
}