线段树分治初步学习&洛谷P5227[AHOI2013]连通图

线段树分治

其实思想说起来是比较简单的，我们把这个题里的所有操作（比如连边删边查询balabala）全部拍到一棵线段树上，然后对着整棵树dfs一下求解答案，顺便把操作做一下，回溯的时候撤销一下即可。虽然有的操作需要以区间形式拍到树上，导致它可能会被拆成两个，但线段树的形态同样保证了操作最多只会被拆分\(log(区间长度)\)次，保障了复杂度。

洛谷P5227[AHOI2013]连通图

传送门

其实就是线段树分治+带撤销并查集，并查集写按秩合并，不能路径压缩（否则会破坏结构，就会撤销出奇怪的效果）
以询问的时间轴为下标建一棵线段树，然后把边存在的时间区间拍到线段树上，用\(vector\)存下来，再对着树一波\(dfs\)记录答案，注意在叶节点不能写\(return\)我沙茶了，不然叶节点如果有操作就会还没撤销就返回了。
md没有板对着敲自己yy着写好难受，码了一天

/*P5227 [AHOI2013]连通图*/  
#include <bits/stdc++.h>
#define N (100000 + 5)
using namespace std;
inline int read() {
	int cnt = 0, f = 1; char c = getchar();
	while (!isdigit(c)) {if (c == '-') f = -f; c = getchar();}
	while (isdigit(c)) {cnt = (cnt << 3) + (cnt << 1) + c - '0'; c = getchar();}
	return cnt * f;
}
int n, m, k, c, x, top;
int fa[N];
struct node {
	int u, v;
}edge[N << 1];
struct node2 {
	int siz, dep, fa;
	node2(int siz_ = 1, int dep_ = 1, int fa_ = 0) : siz(siz_), dep(dep_), fa(fa_){};
}bcj[N << 1], ctrl_Z[N << 1];
int cur[N << 1], top2;
int pre[N << 1];
int get_father(int x) {return x == bcj[x].fa ? x : get_father(bcj[x].fa);}
void merge(int p, int q) {
	int x = get_father(p), y = get_father(q);
	if (x == y) return;
	if (bcj[x].dep > bcj[y].dep) swap(x, y);
	ctrl_Z[++top] = bcj[x];
	ctrl_Z[++top] = bcj[y];
	bcj[x].fa = y, bcj[y].dep = max(bcj[y].dep, bcj[x].dep + 1), bcj[y].siz += bcj[x].siz;
	cur[++top2] = x;
	cur[++top2] = y;
}
struct node3{
	int l, r;
	vector<node> E;
	#define l(p) tree[p].l
	#define r(p) tree[p].r
}tree[N << 2];

void build(int l, int r, int p) {
	l(p) = l, r(p) = r;
	if (l == r) return;
	int mid = (l + r) >> 1;
	build (l, mid, p << 1);
	build (mid + 1, r, p << 1 | 1);
}

void insert(node x, int l, int r, int p) {
	if (l <= l(p) && r >= r(p)) { tree[p].E.push_back(x); return; }
	register int mid = (l(p) + r(p)) >> 1;
	if (l <= mid) insert(x, l, r, p << 1);
	if (r > mid) insert(x, l, r, p << 1 | 1);
}
void dfs_(int p) {
	int tp = top2;
	for (register unsigned int i = 0; i < tree[p].E.size(); i++) {
		node now = tree[p].E[i];
		merge(now.u, now.v);
	}
	if (l(p) == r(p)) {
		int now = bcj[get_father(1)].siz;
		printf(now == n ? "Connected\n" : "Disconnected\n");
	} else dfs_(p << 1), dfs_(p << 1 | 1);
	for (; top2 > tp; --top2, --top) {bcj[cur[top2]] = ctrl_Z[top];}
}

int main() {
	n = read(), m = read();
	for (register int i = 1; i <= n; i++) bcj[i] = node2(1, 1, i);
	for (register int i = 1; i <= m; i++) 
		edge[i].u = read(), edge[i].v = read(), pre[i] = 1;
	k = read();
	build (1, k, 1);
	for (register int i = 1; i <= k; i++) {
		c = read();
		for (register int j = 1; j <= c; j++) {
			x = read();
			if (pre[x] < i) insert(edge[x], pre[x], i - 1, 1);
			pre[x] = i + 1;
		}
	}
	for (register int i = 1; i <= m; i++) {
		if (pre[i] <= k) insert(edge[i], pre[i], k, 1);
	}
	dfs_(1);
	return 0;
}