BZOJ 4025: 二分图

线段树分治+按秩合并的并查集解决加边删边的问题。
一个图是二分图当且仅当点数大于等于二并且不存在奇环。
那么可以用带权并查集维护路径长度，会出现环就是当加入一条边是产生环并且原路径长度为偶数。

#include <bits/stdc++.h>

namespace IO {
	void read() {}
	template<class T, class... T2>
	void read(T &x, T2 &... oth) {
		x = 0; T f = 1; char ch = getchar();
		while (!isdigit(ch)) { if (ch == '-') f = -1; ch = getchar(); }
		while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
		x *= f;
		read(oth...);
	}
}

const int N = 2e5 + 7;

int st[N], top;

struct DSU {
	int fa[N], d[N], dep[N];
	void init(int n) {
		for (int i = 0; i <= n; i++)
			fa[i] = i, d[i] = dep[i] = 0;
	}
	int getfa(int x) {
		while (x != fa[x]) 
			x = fa[x];
		return x;
	}
	int getdis(int x) {
		int dis = 0;
		while (x != fa[x])
			dis ^= d[x], x = fa[x];
		return dis;
	}
	bool merge(int x, int y) {
		int D = getdis(x) ^ getdis(y) ^ 1;
		x = getfa(x), y = getfa(y);
		if (x == y) return D == 0;
		if (dep[x] < dep[y]) 
			std::swap(x, y);
		if (dep[x] == dep[y])
			dep[x]++, st[++top] = -x;
		fa[y] = x; d[y] = D; st[++top] = y;
		return 1;
	}
	void del(int tp) {
		while (top > tp) {
			int x = st[top--];
			if (x < 0)
				dep[-x]--;
			else 
				fa[x] = x, d[x] = 0;
		}
	}
} dsu;

int n, m, T;
struct P {
	int u, v, x, y;
} p[N];
int ans[N];

void solve(int l, int r, const std::vector<int> &vec) {
	int tp = top;
	int mid = l + r >> 1;
	std::vector<int> L, R;
	for (int i = 0, sz = vec.size(); i < sz; i++) {
		int id = vec[i];
		if (p[id].x <= l && r <= p[id].y) {
			if (!dsu.merge(p[id].u, p[id].v)) {
				for (int j = l; j <= r; j++)
					ans[j] = 1;
				dsu.del(tp);
				return;
			}
		} else {
			if (p[id].x <= mid)
				L.push_back(id);
			if (p[id].y > mid)
				R.push_back(id);
		}
	}
	if (l == r) {
		dsu.del(tp);
		return;
	}
	solve(l, mid, L);
	solve(mid + 1, r, R);
	dsu.del(tp);
}

int main() {
	IO::read(n, m, T);
	dsu.init(n);
	std::vector<int> vec;
	for (int i = 1; i <= m; i++)
		IO::read(p[i].u, p[i].v, p[i].x, p[i].y), p[i].x++, vec.push_back(i);
	solve(1, T, vec);
	for (int i = 1; i <= T; i++)
		puts(ans[i] ? "No" : "Yes");
	return 0;
}