BZOJ 3514: Codechef MARCH14 GERALD07加强版 (LCT维护最大生成树+主席树)

题意

给出$n$个点,$m$条边.多次询问,求编号在$[l,r]$内的边形成的联通块的数量,强制在线.

分析

• $LCT$维护动态最大生成树，先将每条边依次加进去，若形成环就断掉最早加进去(编号最小)的边，然后记录$early[]$数组，表示第i条边弹掉了哪条边，若没有弹出边，$early[i]=0$
• 然后每个询问的答案就是用n减掉[l,r]区间内early值小于l的边的数量,可以用主席树来维护
• 正确性证明:因为是维护的最大生成树,而$early[i]$又是最大生成树中的最小边,那么$(early[i],i)$这些边一定不能让$i$的两个端点联通.所以说对于$r>=i$且$l>early[i]$的询问,如果按编号从小到大加入边,那么在$i$之前它的两个端点一定还不联通,加入$i$这条边让它们合并成为一个联通块,答案就减一.所以说只要求出所有$early$在主席树上维护就行了.
• 注意有自环(良心样例有自环)

CODE

#include <cctype>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
char cb[1<<15],*cs=cb,*ct=cb;
#define getc() (cs==ct&&(ct=(cs=cb)+fread(cb,1,1<<15,stdin),cs==ct)?0:*cs++)
template<class T>inline void read(T &res) {
    char ch; for(;!isdigit(ch=getc()););
    for(res=ch-'0';isdigit(ch=getc());res=res*10+ch-'0');
}
const int MAXN = 200005;
const int MAXM = MAXN*20;
int n;
namespace LCT {
	const int N = 400005;
	#define ls ch[x][0]
	#define rs ch[x][1]
	int ch[N][2], fa[N], mn[N]; bool rev[N];
	inline bool isr(int x) { return ch[fa[x]][0] != x && ch[fa[x]][1] != x; }
	inline bool get(int x) { return ch[fa[x]][1] == x; }
	inline int Max(int x, int y) { if(x <= n) return y; if(y <= n) return x; return min(x, y); }
	inline void upd(int x) { mn[x] = Max(x, Max(mn[ls], mn[rs]));}
	inline void mt(int x) { if(rev[x]) rev[x]^=1, rev[ls]^=1, rev[rs]^=1, swap(ls, rs); }
	void mtpath(int x) { if(!isr(x)) mtpath(fa[x]); mt(x); }
	inline void rot(int x) {
		int y = fa[x], z = fa[y];
		bool l = get(x), r = l^1;
		if(!isr(y)) ch[z][get(y)] = x;
		fa[ch[x][r]] = y, fa[y] = x, fa[x] = z;
		ch[y][l] = ch[x][r], ch[x][r] = y;
		upd(y), upd(x);
	}
	inline void splay(int x) {
		mtpath(x);
		for(; !isr(x); rot(x))
			if(!isr(fa[x])) rot(get(x) == get(fa[x]) ? fa[x] : x);
	}
	inline int access(int x) { int y = 0;
		for(; x; x = fa[y=x]) splay(x), ch[x][1] = y, upd(x);
		return y;
	}
	inline void bert(int x) { access(x), splay(x), rev[x]^=1; }
	inline int sert(int x) { access(x), splay(x); for(; ls; x=ls); return x; }
	inline bool judge(int x, int y) { bert(x); return sert(y) == x; }
	inline void link(int x, int y) { bert(x); fa[x] = y; }
	inline void split(int x, int y) { bert(x); access(y), splay(y); }
	inline void cut(int x, int y) { split(x, y); ch[y][0] = fa[x] = 0; upd(y); }
	inline int findmn(int x, int y) { split(x, y); return mn[y]; }
	#undef ls
	#undef rs
}
int m, q, type, pre[MAXN], u[MAXN], v[MAXN];
int rt[MAXN], sz, sum[MAXM], ch[MAXM][2];
void modify(int &i, int p, int l, int r, int x) {
	if(!i) i = ++sz;
	sum[i] = sum[p] + 1;
	if(l == r) return;
	int mid = (l + r) >> 1;
	if(x <= mid) ch[i][1] = ch[p][1], modify(ch[i][0], ch[p][0], l, mid, x);
	else ch[i][0] = ch[p][0], modify(ch[i][1], ch[p][1], mid+1, r, x);
}
int query(int i, int j, int l, int r, int ql, int qr) {
	if(l == ql && r == qr) return sum[j] - sum[i];
	int mid = (l + r) >> 1;
	if(qr <= mid) return query(ch[i][0], ch[j][0], l, mid, ql, qr);
	else if(ql > mid) return query(ch[i][1], ch[j][1], mid+1, r, ql, qr);
	return query(ch[i][0], ch[j][0], l, mid, ql, mid) + query(ch[i][1], ch[j][1], mid+1, r, mid+1, qr);
}
int main () {
	read(n), read(m), read(q), read(type);
	for(int i = 1; i <= m; ++i) {
		read(u[i]), read(v[i]);
		if(u[i] == v[i]) rt[i] = rt[i-1];
		else {
			if(LCT::judge(u[i], v[i])) {
				int p = LCT::findmn(u[i], v[i]);
				pre[i] = p-n;
				LCT::cut(u[pre[i]], p);
				LCT::cut(v[pre[i]], p);
			}
			LCT::link(u[i], n+i);
			LCT::link(v[i], n+i);
			modify(rt[i], rt[i-1], 0, m, pre[i]);
		}
	}
	int lastans = 0, x, y;
	while(q--) {
		read(x), read(y);
		if(type) x ^= lastans, y ^= lastans;
		printf("%d\n", lastans = (n - query(rt[x-1], rt[y], 0, m, 0, x-1)));
	}
}