[COCI 2023/2024 #1] Mostovi 题解

前言

题目链接：洛谷。

题目分析

首先可以确定的是需要枚举断边，所以我们希望两次枚举之间能有些关联。不难想到类树形 DP 的套路，建 DFS 树，只不过这题除了讨论和父亲之间的边，还要考虑返租边。以下钦定以 \(1\) 为树根。

树边

先从简单的树边开始考虑。考虑不经过 \(u\) 和 \(u\) 的父亲 \(v\)，对答案是否产生贡献。为方便实现，记录的是操作后整张图依旧联通的边数。

分为两种情况讨论。

\(v\) 为树根。
1. \(v\) 仅有 \(u\) 一个儿子。操作后整张图依旧联通。
2. \(v\) 还有另一个儿子 \(yzh\)。整张图联通当且仅当 \(u\) 是叶子结点，不然 \(u\) 的子孙和 \(yzh\) 就断开了。
3. \(v\) 还有多个儿子。无论怎样，由于不能经过 \(v\)，剩下来的孩子都互不连通。
\(v\) 为内部节点。
1. 最好的情况，\(v\) 的所有子树都能向上连到 \(v\) 的祖先，\(u\) 的所有子树同样能连到 \(v\) 的祖先，操作后整张图依旧连通。
2. \(u\) 是叶子结点，并且 \(v\) 的其他子树都能连到 \(v\) 的祖先，操作后整张图依旧连通。
3. 其他情况操作后图均不连通。

可以通过下图辅助理解。

情况 1

情况 2

具体实现时，按照上述方法模拟即可。具体如下：

用一次 dfs 记录 \(1\) 的儿子数量和 \(u\) 儿子数量，再用一次 dfs 统计答案即可。
用一次 dfs 求出以 \(u\) 为根的子树通过一条返租边能到达的最深结点，和次深结点。由于返租边的另一边只能是祖先结点，所以可以用深度来代替返租边另一边的结点。再用另一次 dfs 按照如下顺序求解。
1. \(u\) 有一个结点不能连到 \(u\) 的祖先，那么不管如何，删去 \(u\) 这个点一定会导致图不连通。
2. \(u\) 的某一个子树只有一条返租边连向 \(u\) 的祖先 \(xym\)，那么如果存在 \(u \rightarrow xym\) 这条边，删去后图不连通，可以用一个数据结构维护不能删去的点（实际记录深度即可）。
3. 考虑 \(u \rightarrow v\) 这条树边，如果 \(v \neq xym\) 并且 \(v\) 的子树中没有不能连向 \(v\) 的祖先的子树，或者仅有一个并且其为身为叶子结点的 \(u\)，删去后图依旧联通。

非树边

考虑完树边，再来看看相对复杂的非树边 \(u \rightarrow yzh\)，易知 \(yzh\) 一定是 \(u\) 的祖先。

经过我们观察，删去非树边比删去树边多出了 \(u \rightarrow \cdots \rightarrow yzh\) 这一条链上的部分，即 \(yzh\) 的子树减去 \(u\) 的子树，如下图蓝色部分。

yzh 520

所以我们考虑图的连通性的时候就要算上它们，其他和以上讨论类似。

\(yzh\) 是树根，按照以上特判即可。
红色部分不连通，删去一定不连通。
\(u\) 有一棵子树既不能连到红色部分，又不能连到蓝色部分，删去一定不连通。
\(u\) 为叶子结点，或者所有子树能和红色或蓝色其中一个部分连通，当且仅当红色和蓝色部分连通，删去后连通。
有一棵子树既和红色部分连通又和蓝色部分连通，删去图依旧联通。
其他情况删去一定不连通。

~~很抽象是吧~~，这道题目考验我们强大的逻辑推理能力，做到情况不漏，实现细心。所以来讲讲具体实现。

按照树边特判即可。
记录 \(yzh\) 有多少棵子树不与 \(yzh\) 祖先连通，如果大于一个，删去不连通。
这种情况相当于有一棵子树不能练到 \(u\) 的祖先结点，这个同上解决。也一样记录特殊情况，即 \(u\) 的某一个子树只有一条返租边连向 \(u\) 的祖先 \(xym\)，那么不能删掉 \(xym\) 这个点。
我们想知道蓝色部分能不能连到红色部分，发现可以用倍增解决出蓝色部分能连出的最浅深度。这个倍增不能包含 \(u\) 子树的信息。
相当于 \(yzh\) 处在 \(u\) 连出的某两条边之间，那么记录 \(u\) 子树中能连出的最浅深度，这个要比 \(yzh\) 浅，然后记录连到 \(u\) 祖先最深的深度，这个要比 \(yzh\) 深。前者第一次 dfs 可以预处理出，后者发现需要将子树里的返租边，但是反不到 \(u\) 的返租边删除，即只保留另一端深度小于 \(u\) 的返租边。这个可以用 dfs 序加线段树或者左偏树，但是最方便的做法当然是愉快地启发式合并啦。

代码（已略去快读快写）

讲了这么多，实际代码需要格外的小心，注意细节。令 \(n\) 和 \(m\) 同阶，那么整体时间复杂度是 \(\Theta(n \log ^ 2 n)\) 或者 \(\Theta(n \log n)\) 的。

左偏树

点击查看代码

//#pragma GCC optimize(3)
//#pragma GCC optimize("Ofast", "inline", "-ffast-math")
//#pragma GCC target("avx", "sse2", "sse3", "sse4", "mmx")
#include <iostream>
#include <cstdio>
#define debug(a) cerr << "Line: " << __LINE__ << " " << #a << endl
#define print(a) cerr << #a << "=" << (a) << endl
#define file(a) freopen(#a".in", "r", stdin), freopen(#a".out", "w", stdout)
#define main Main(); signed main(){ return ios::sync_with_stdio(0), cin.tie(0), Main(); } signed Main
using namespace std;

#include <algorithm>
#include <vector>
#include <set>

const int inf = 0x3f3f3f3f;
const int  N  = 100010;
const int  M  = 300010;
const int lgN = __lg(N) + 2;

struct Graph{
	struct node{
		int to, nxt;
	} edge[M << 1];
	int eid, head[N];
	inline void add(int u, int v){
		edge[++eid] = {v, head[u]};
		head[u] = eid;
	}
	inline node & operator [] (const int x){
		return edge[x];
	}
} xym;

struct node{
	int fi, se;
	node (int fi = inf, int se = inf): fi(fi), se(se) {}
	inline node operator + (const int val) const {
		node res(fi, se);
		if (val < fi) res.se = res.fi, res.fi = val;
		else if (fi < val && val < se) res.se = val;
		return res;
	}
	inline node operator + (const node & o) const {
		return *this + o.fi + o.se;
	}
	inline node & operator += (const node & o) {
		return *this = *this + o;
	}
};

int n, m, ans;

int dpt[N], yzh[N][lgN];
int L[N], R[N], timer;
vector<int> son[N];
node f[N][lgN], pos[N], s[N];
int sum[N], who[N];

void dfs(int now){
	L[now] = ++timer, dpt[now] = dpt[yzh[now][0]] + 1;
	vector<node> l, r; int cnt = 0;
	for (int i = xym.head[now]; i; i = xym[i].nxt){
		int to = xym[i].to;
		if (to == yzh[now][0]) continue;
		if (dpt[to]) pos[now] += dpt[to];
		else {
			son[now].push_back(to), yzh[to][0] = now, ++cnt;
			dfs(to), s[now] += s[to], l.push_back(s[to]), r.push_back(s[to]);
			if (s[to].fi >= dpt[now]) ++sum[now], who[now] = to;
		}
	}
	for (int i = 1; i < cnt; ++i)      l[i] += l[i - 1];
	for (int i = cnt - 2; i >= 0; --i) r[i] += r[i + 1];
	for (int i = 0; i < cnt; ++i){
		int to = son[now][i];
		if (i > 0) f[to][0] += l[i - 1];
		if (i + 1 < cnt) f[to][0] += r[i + 1];
		f[to][0] += pos[now];
	}
	R[now] = timer, s[now] += pos[now];
}

int root[N], tot;
int deleted[M], dtop;
int lson[M], rson[M], val[M];
int dist[M];

int newNode(int val){
	int res = dtop ? deleted[dtop--] : ++tot;
	lson[res] = rson[res] = dist[res] = 0;
	::val[res] = val;
	return res;
}

int combine(int x, int y){
	if (!x || !y) return x | y;
	if (val[x] < val[y]) swap(x, y);
	rson[x] = combine(rson[x], y);
	if (dist[rson[x]] > dist[lson[x]]) swap(lson[x], rson[x]);
	return dist[x] = dist[rson[x]] + 1, x;
}

void redfs(int now){
	set<int> not_ok;
	set<pair<int, int> > stt;
	bool flag = false;
	for (const auto& to: son[now]){
		redfs(to);
		while (root[to] && val[root[to]] >= dpt[now]){
			deleted[++dtop] = root[to];
			root[to] = combine(lson[root[to]], rson[root[to]]);
		}
		if (root[to] && val[root[to]] != s[to].fi)
			stt.insert({val[root[to]], s[to].fi});
		root[now] = combine(root[now], root[to]);
		if (s[to].fi >= dpt[now]) flag = true;
		else if (s[to].se >= dpt[now]) not_ok.insert(s[to].fi);
	}
	vector<pair<int, int> > l(stt.begin(), stt.end()); stt.clear();
	int tot = l.size();
	for (int i = tot - 2; i >= 0; --i)
		l[i].second = min(l[i].second, l[i + 1].second);
	for (int i = xym.head[now]; i; i = xym[i].nxt){
		int to = xym[i].to;
		if (dpt[to] > dpt[now]) continue;
		root[now] = combine(root[now], newNode(dpt[to]));
		if (to == 1){
			if (sum[to] != 1){
				if (to == yzh[now][0] && son[now].size() == 0u && sum[to] == 2) ++ans;
			} else {
				if (to == yzh[now][0] && son[now].size() != 1u) continue;
				if (to != yzh[now][0] && (flag || not_ok.count(1))) continue;
				++ans;
			}
			continue;
		}
		if (flag) continue;
		if (to != yzh[now][0]){
			node sum; int u = now;
			for (int k = lgN - 1; k >= 0; --k)
			if (yzh[u][k] && dpt[yzh[u][k]] > dpt[to])
				sum += f[u][k], u = yzh[u][k];
			if (sum.fi >= dpt[to]){
				vector<pair<int, int> >::iterator it = 
					upper_bound(l.begin(), l.end(), pair<int, int>{dpt[to], inf});
				if (it == l.end() || it -> second >= dpt[to]) continue;
			}
		}
		if (sum[to] > 1) continue;
		if (sum[to] != 0 && (L[now] < L[who[to]] || R[who[to]] < L[now])) continue;
		if (not_ok.count(dpt[to])) continue;
		++ans;
	}
}

signed main(){
	read(n, m);
	for (int i = 1, u, v; i <= m; ++i) read(u, v), xym.add(u, v), xym.add(v, u);
	dfs(1);
	for (int k = 1; k < lgN; ++k)
	for (int i = 1; i <= n; ++i){
		yzh[i][k] = yzh[yzh[i][k - 1]][k - 1];
		f[i][k] = f[i][k - 1] + f[yzh[i][k - 1]][k - 1];
	}
	redfs(1), write(m - ans);
	return 0;
}

启发式合并

点击查看代码

//#pragma GCC optimize(3)
//#pragma GCC optimize("Ofast", "inline", "-ffast-math")
//#pragma GCC target("avx", "sse2", "sse3", "sse4", "mmx")
#include <iostream>
#include <cstdio>
#define debug(a) cerr << "Line: " << __LINE__ << " " << #a << endl
#define print(a) cerr << #a << "=" << (a) << endl
#define file(a) freopen(#a".in", "r", stdin), freopen(#a".out", "w", stdout)
#define main Main(); signed main(){ return ios::sync_with_stdio(0), cin.tie(0), Main(); } signed Main
using namespace std;

#include <algorithm>
#include <vector>
#include <set>

const int inf = 0x3f3f3f3f;
const int  N  = 100010;
const int  M  = 300010;
const int lgN = __lg(N) + 2;

struct Graph{
	struct node{
		int to, nxt;
	} edge[M << 1];
	int eid, head[N];
	inline void add(int u, int v){
		edge[++eid] = {v, head[u]};
		head[u] = eid;
	}
	inline node & operator [] (const int x){
		return edge[x];
	}
} xym;

struct node{
	int fi, se;
	node (int fi = inf, int se = inf): fi(fi), se(se) {}
	inline node operator + (const int val) const {
		node res(fi, se);
		if (val < fi) res.se = res.fi, res.fi = val;
		else if (fi < val && val < se) res.se = val;
		return res;
	}
	inline node operator + (const node & o) const {
		return *this + o.fi + o.se;
	}
	inline node & operator += (const node & o) {
		return *this = *this + o;
	}
};

int n, m, ans;

int dpt[N], yzh[N][lgN];
int L[N], R[N], timer;
vector<int> son[N];
node f[N][lgN], pos[N], s[N];
int sum[N], who[N];

void dfs(int now){
	L[now] = ++timer, dpt[now] = dpt[yzh[now][0]] + 1;
	vector<node> l, r; int cnt = 0;
	for (int i = xym.head[now]; i; i = xym[i].nxt){
		int to = xym[i].to;
		if (to == yzh[now][0]) continue;
		if (dpt[to]) pos[now] += dpt[to];
		else {
			son[now].push_back(to), yzh[to][0] = now, ++cnt;
			dfs(to), s[now] += s[to], l.push_back(s[to]), r.push_back(s[to]);
			if (s[to].fi >= dpt[now]) ++sum[now], who[now] = to;
		}
	}
	for (int i = 1; i < cnt; ++i)      l[i] += l[i - 1];
	for (int i = cnt - 2; i >= 0; --i) r[i] += r[i + 1];
	for (int i = 0; i < cnt; ++i){
		int to = son[now][i];
		if (i > 0) f[to][0] += l[i - 1];
		if (i + 1 < cnt) f[to][0] += r[i + 1];
		f[to][0] += pos[now];
	}
	R[now] = timer, s[now] += pos[now];
}

set<int> st[N];

inline void merge(set<int> &a, set<int>& b){
	if (a.size() < b.size()) a.swap(b);
	for (const auto& x: b) a.insert(x);
	b.clear();
}

void redfs(int now){
	set<int> not_ok;
	set<pair<int, int> > stt;
	bool flag = false;
	for (const auto& to: son[now]){
		redfs(to);
		while (!st[to].empty() && *st[to].rbegin() >= dpt[now])
			st[to].erase(prev(st[to].end()));
		if (st[to].size() > 1u) stt.insert({*st[to].rbegin(), *st[to].begin()});
		merge(st[now], st[to]);
		if (s[to].fi >= dpt[now]) flag = true;
		else if (s[to].se >= dpt[now]) not_ok.insert(s[to].fi);
	}
	vector<pair<int, int> > l(stt.begin(), stt.end()); stt.clear();
	int tot = l.size();
	for (int i = tot - 2; i >= 0; --i)
		l[i].second = min(l[i].second, l[i + 1].second);
	for (int i = xym.head[now]; i; i = xym[i].nxt){
		int to = xym[i].to;
		if (dpt[to] > dpt[now]) continue;
		st[now].insert(dpt[to]);
		if (to == 1){
			if (sum[to] != 1){
				if (to == yzh[now][0] && son[now].size() == 0u && sum[to] == 2) ++ans;
			} else {
				if (to == yzh[now][0] && son[now].size() != 1u) continue;
				if (to != yzh[now][0] && (flag || not_ok.count(1))) continue;
				++ans;
			}
			continue;
		}
		if (flag) continue;
		if (to != yzh[now][0]){
			node sum; int u = now;
			for (int k = lgN - 1; k >= 0; --k)
			if (yzh[u][k] && dpt[yzh[u][k]] > dpt[to])
				sum += f[u][k], u = yzh[u][k];
			if (sum.fi >= dpt[to]){
				vector<pair<int, int> >::iterator it = 
					upper_bound(l.begin(), l.end(), pair<int, int>{dpt[to], inf});
				if (it == l.end() || it -> second >= dpt[to]) continue;
			}
		}
		if (sum[to] > 1) continue;
		if (sum[to] != 0 && (L[now] < L[who[to]] || R[who[to]] < L[now])) continue;
		if (not_ok.count(dpt[to])) continue;
		++ans;
	}
}

signed main(){
	read(n, m);
	for (int i = 1, u, v; i <= m; ++i) read(u, v), xym.add(u, v), xym.add(v, u);
	dfs(1);
	for (int k = 1; k < lgN; ++k)
	for (int i = 1; i <= n; ++i){
		yzh[i][k] = yzh[yzh[i][k - 1]][k - 1];
		f[i][k] = f[i][k - 1] + f[yzh[i][k - 1]][k - 1];
	}
	redfs(1), write(m - ans);
	return 0;
}

dfs 序加线段树

点击查看代码

//#pragma GCC optimize(3)
//#pragma GCC optimize("Ofast", "inline", "-ffast-math")
//#pragma GCC target("avx", "sse2", "sse3", "sse4", "mmx")
#include <iostream>
#include <cstdio>
#define debug(a) cerr << "Line: " << __LINE__ << " " << #a << endl
#define print(a) cerr << #a << "=" << (a) << endl
#define file(a) freopen(#a".in", "r", stdin), freopen(#a".out", "w", stdout)
#define main Main(); signed main(){ return ios::sync_with_stdio(0), cin.tie(0), Main(); } signed Main
using namespace std;

#include <unordered_set>
#include <algorithm>
#include <vector>
#include <set>

const int inf = 0x3f3f3f3f;
const int  N  = 100010;
const int  M  = 300010;
const int lgN = __lg(N) + 2;

struct Graph{
	struct node{
		int to, nxt;
	} edge[M << 1];
	int eid, head[N];
	inline void add(const int &u, const int &v){
		edge[++eid] = {v, head[u]};
		head[u] = eid;
	}
	inline node & operator [] (const int x){
		return edge[x];
	}
} xym;

struct node{
	int fi, se;
	inline node (int fi = inf, int se = inf): fi(fi), se(se) {}
	inline void merge(const node & o){
		this -> merge(o.fi);
		this -> merge(o.se);
	}
	inline void merge(const int val){
		if (val < fi) se = fi, fi = val;
		else if (fi < val && val < se) se = val;
	}
};

int n, m, ans;
int real_lgN;

int dpt[N], yzh[N][lgN];
int L[N], R[N], timer;
vector<int> son[N], rev[N];
node f[N][lgN], pos[N], s[N];
int sum[N], who[N];

void dfs(int now){
	L[now] = ++timer, dpt[now] = dpt[yzh[now][0]] + 1;
	vector<node> l, r; int cnt = 0;
	for (register int i = xym.head[now]; i; i = xym[i].nxt){
		int to = xym[i].to;
		if (to == yzh[now][0]) continue;
		if (dpt[to]){
			pos[now].merge(dpt[to]);
			if (dpt[to] <= dpt[now]) rev[now].emplace_back(to);
		} else {
			son[now].emplace_back(to), yzh[to][0] = now;
			rev[to].emplace_back(now);
			dfs(to), s[now].merge(s[to]), l.emplace_back(s[to]);
			if (s[to].fi >= dpt[now]) ++sum[now], who[now] = to;
		}
	}
	r = l, cnt = son[now].size();
	for (register int i = cnt - 2; i >= 0; --i) r[i].merge(r[i + 1]);
	for (register int i = 0; i < cnt; ++i){
		register int to = son[now][i];
		(i > 0) && (f[to][0].merge(l[i - 1]), l[i].merge(l[i - 1]), yzh_i_love_you);
		(i + 1 < cnt) && (f[to][0].merge(r[i + 1]), yzh_i_love_you);
		f[to][0].merge(pos[now]);
	}
	R[now] = timer, s[now].merge(pos[now]);
}

int iL[N], iR[N], itimer;
int val[M];
void build(int now){
	iL[now] = itimer + 1;
	for (const auto& to: rev[now]) val[++itimer] = dpt[to];
	for (const auto& to: son[now]) build(to);
	iR[now] = itimer;
}

struct Segment_Tree{
	#define lson (idx << 1    )
	#define rson (idx << 1 | 1)
	
	struct Info{
		int maxx, maxpos;
		int minn, minpos;
		inline Info operator + (const Info& o) const {
			if (maxx == -inf) return o;
			if (o.maxx == -inf) return *this;
			return {maxx > o.maxx ? maxx : o.maxx, maxx > o.maxx ? maxpos : o.maxpos,
					minn < o.minn ? minn : o.minn, minn < o.minn ? minpos : o.minpos};
		}
	};
	
	struct node{
		int l, r;
		Info info;
	} tree[M << 2];
	
	inline void pushup(int idx){
		tree[idx].info = tree[lson].info + tree[rson].info;
	}
	
	void build(int idx, int l, int r){
		tree[idx] = {l, r, {-inf, -1, inf, -1}};
		if (l == r) return tree[idx].info = {val[l], l, val[l], l}, void();
		int mid = (l + r) >> 1;
		build(lson, l, mid), build(rson, mid + 1, r);
		pushup(idx);
	}
	
	Info query(int idx, int l, int r){
		if (tree[idx].l > r || tree[idx].r < l) return {-inf, -1, inf, -1};
		if (l <= tree[idx].l && tree[idx].r <= r) return tree[idx].info;
		return query(lson, l, r) + query(rson, l, r);
	}
	
	void erase(int idx, int p){
		if (tree[idx].l > p || tree[idx].r < p) return;
		if (tree[idx].l == tree[idx].r) return tree[idx].info = {-inf, -1, inf, -1}, void();
		erase(lson, p), erase(rson, p), pushup(idx);
	}
	
	#undef lson
	#undef rson
} tree;

void redfs(int now){
	if (iL[now] > iR[now]) return;
	unordered_set<int> not_ok;
	vector<pair<int, int> > l;
	bool flag = false;
	for (const auto& to: son[now]){
		redfs(to);
		while (iL[to] <= iR[to]){
			Segment_Tree::Info res = tree.query(1, iL[to], iR[to]);
			if (res.minn == inf || res.maxx == -inf) break;
			if (res.maxx < dpt[now]) break;
			tree.erase(1, res.maxpos);
		}
		if (iL[to] <= iR[to]){
			Segment_Tree::Info res = tree.query(1, iL[to], iR[to]);
			if (res.minn != inf && res.maxx != -inf && res.minn != res.maxx)
				l.emplace_back(res.maxx, res.minn);
		}
		if (s[to].fi >= dpt[now]) flag = true;
		else if (s[to].se >= dpt[now]) not_ok.insert(s[to].fi);
	}
	sort(l.begin(), l.end());
	int tot = l.size();
	for (register int i = tot - 2; i >= 0; --i)
		l[i].second = Fast::min(l[i].second, l[i + 1].second);
	for (const auto& to: rev[now]){
		if (to == 1){
			if (sum[to] != 1){
				if (to == yzh[now][0] && son[now].size() == 0u && sum[to] == 2) ++ans;
			} else {
				if (to == yzh[now][0] && son[now].size() != 1u) continue;
				if (to != yzh[now][0] && (flag || not_ok.count(1))) continue;
				++ans;
			}
			continue;
		}
		if (flag) continue;
		if (to != yzh[now][0]){
			node sum; register int u = now;
			for (register int k = real_lgN - 1; k >= 0; --k)
			if (yzh[u][k] && dpt[yzh[u][k]] > dpt[to])
				sum.merge(f[u][k]), u = yzh[u][k];
			if (sum.fi >= dpt[to]){
				vector<pair<int, int> >::iterator it = 
					upper_bound(l.begin(), l.end(), pair<int, int>{dpt[to], inf});
				if (it == l.end() || it -> second >= dpt[to]) continue;
			}
		}
		if (sum[to] > 1 || not_ok.count(dpt[to])) continue;
		if (sum[to] != 0 && (L[now] < L[who[to]] || R[who[to]] < L[now])) continue;
		++ans;
	}
}

signed main(){
	read(n, m), real_lgN = __lg(n) + 2;
	for (int i = 1, u, v; i <= m; ++i) read(u, v), xym.add(u, v), xym.add(v, u);
	dfs(1);
	for (register int k = 1; k < real_lgN; ++k)
	for (register int i = 1; i <= n; ++i){
		yzh[i][k] = yzh[yzh[i][k - 1]][k - 1];
		f[i][k] = f[i][k - 1];
		if (yzh[i][k - 1]) f[i][k].merge(f[yzh[i][k - 1]][k - 1]);
	}
	build(1), tree.build(1, 1, itimer), redfs(1), write(m - ans);
	return 0;
}

线段树合并

~~补题降智以为 \(\Theta(n \log n)\) 的线段树合并是 \(\Theta(\log n)\) 的，总的时间复杂度是 \(\Theta(n ^ 2 \log n)\) 的，喜提 TLE。但是还是放出来吧，写都写了。~~

UPDATE on 2024.4.7

~~好吧，写题解更降智。~~ 维护一个权值线段树，然后向上合并。你干嘛要删除啊，每次查询 \([1, dpt[now])\) 里的最大值就行了，这样就不会超时。

AC code

点击查看代码

//#pragma GCC optimize(3)
//#pragma GCC optimize("Ofast", "inline", "-ffast-math")
//#pragma GCC target("avx", "sse2", "sse3", "sse4", "mmx")
#include <iostream>
#include <cstdio>
#define debug(a) cerr << "Line: " << __LINE__ << " " << #a << endl
#define print(a) cerr << #a << "=" << (a) << endl
#define file(a) freopen(#a".in", "r", stdin), freopen(#a".out", "w", stdout)
#define main Main(); signed main(){ return ios::sync_with_stdio(0), cin.tie(0), Main(); } signed Main
using namespace std;

#include <unordered_set>
#include <algorithm>
#include <vector>
#include <set>

const int inf = 0x3f3f3f3f;
const int  N  = 100010;
const int  M  = 300010;
const int lgN = __lg(N) + 2;

struct Graph{
	struct node{
		int to, nxt;
	} edge[M << 1];
	int eid, head[N];
	inline void add(const int &u, const int &v){
		edge[++eid] = {v, head[u]};
		head[u] = eid;
	}
	inline node & operator [] (const int x){
		return edge[x];
	}
} xym;

struct node{
	int fi, se;
	inline node (int fi = inf, int se = inf): fi(fi), se(se) {}
	inline void merge(const node & o){
		this -> merge(o.fi);
		this -> merge(o.se);
	}
	inline void merge(const int val){
		if (val < fi) se = fi, fi = val;
		else if (fi < val && val < se) se = val;
	}
};

int n, m, ans;
int real_lgN;

int dpt[N], yzh[N][lgN];
int L[N], R[N], timer;
vector<int> son[N], rev[N];
node f[N][lgN], pos[N], s[N];
int sum[N], who[N];

void dfs(int now){
	L[now] = ++timer, dpt[now] = dpt[yzh[now][0]] + 1;
	vector<node> l, r; int cnt = 0;
	for (register int i = xym.head[now]; i; i = xym[i].nxt){
		int to = xym[i].to;
		if (to == yzh[now][0]) continue;
		if (dpt[to]){
			pos[now].merge(dpt[to]);
			if (dpt[to] <= dpt[now]) rev[now].emplace_back(to);
		} else {
			son[now].emplace_back(to), yzh[to][0] = now;
			rev[to].emplace_back(now);
			dfs(to), s[now].merge(s[to]), l.emplace_back(s[to]);
			if (s[to].fi >= dpt[now]) ++sum[now], who[now] = to;
		}
	}
	r = l, cnt = son[now].size();
	for (register int i = cnt - 2; i >= 0; --i) r[i].merge(r[i + 1]);
	for (register int i = 0; i < cnt; ++i){
		register int to = son[now][i];
		(i > 0) && (f[to][0].merge(l[i - 1]), l[i].merge(l[i - 1]), yzh_i_love_you);
		(i + 1 < cnt) && (f[to][0].merge(r[i + 1]), yzh_i_love_you);
		f[to][0].merge(pos[now]);
	}
	R[now] = timer, s[now].merge(pos[now]);
}

int root[N];
struct Combinable_Segment_Tree{
	struct node{
		int lson, rson;
		int maxx;
	} tree[N << 5];
	int tot;
	
	int deleted[N << 4], dtop;
	
	int NewNode(){
		int now = dtop ? deleted[dtop--] : ++tot;
		tree[now] = {0, 0, -inf};
		return now;
	}
	
	inline node & operator [] (const int x){
		return tree[x];
	}
	
	inline void pushup(int idx){
		if (tree[idx].lson && tree[idx].rson){
			tree[idx].maxx = Fast::max(tree[tree[idx].lson].maxx, tree[tree[idx].rson].maxx);
		} else if (tree[idx].lson){
			tree[idx].maxx = tree[tree[idx].lson].maxx;
		} else if (tree[idx].rson){
			tree[idx].maxx = tree[tree[idx].rson].maxx;
		} else {
			tree[idx].maxx = -inf;
		}
	}
	
	void insert(int &idx, int trl, int trr, int p){
		if (trl > p || trr < p) return;
		if (!idx) idx = NewNode();
		if (trl == trr) return tree[idx].maxx = p, void();
		int mid = (trl + trr) >> 1;
		insert(tree[idx].lson, trl, mid, p);
		insert(tree[idx].rson, mid + 1, trr, p);
		pushup(idx);
	}
	
	int combine(int idx, int oidx, int trl, int trr){
		if (!idx || !oidx) return idx | oidx;
		if (trl == trr) return tree[idx].maxx = tree[oidx].maxx, idx;
		int mid = (trl + trr) >> 1;
		tree[idx].lson = combine(tree[idx].lson, tree[oidx].lson, trl, mid);
		tree[idx].rson = combine(tree[idx].rson, tree[oidx].rson, mid + 1, trr);
		return pushup(idx), idx;
	}
	
	int query(int idx, int trl, int trr, int l, int r){
		if (!idx || trl > r || trr < l) return -inf;
		if (l <= trl && trr <= r) return tree[idx].maxx;
		int mid = (trl + trr) >> 1;
		return max(query(tree[idx].lson, trl, mid, l, r), query(tree[idx].rson, mid + 1, trr, l, r));
	}
} tree;

void redfs(int now){
	unordered_set<int> not_ok;
	vector<pair<int, int> > l;
	bool flag = false;
	for (const auto& to: son[now]){
		redfs(to);
		if (root[to] && dpt[now] != 1){
			int maxx = tree.query(root[to], 1, n, 1, dpt[now] - 1);
			if (maxx != -inf && s[to].fi != maxx && s[to].fi != -inf)
				l.emplace_back(maxx, s[to].fi);
		}
		root[now] = tree.combine(root[now], root[to], 1, n);
		if (s[to].fi >= dpt[now]) flag = true;
		else if (s[to].se >= dpt[now]) not_ok.insert(s[to].fi);
	}
	sort(l.begin(), l.end());
	int tot = l.size();
	for (register int i = tot - 2; i >= 0; --i)
		l[i].second = Fast::min(l[i].second, l[i + 1].second);
	for (const auto& to: rev[now]){
		tree.insert(root[now], 1, n, dpt[to]);
		if (to == 1){
			if (sum[to] != 1){
				if (to == yzh[now][0] && son[now].size() == 0u && sum[to] == 2) ++ans;
			} else {
				if (to == yzh[now][0] && son[now].size() != 1u) continue;
				if (to != yzh[now][0] && (flag || not_ok.count(1))) continue;
				++ans;
			}
			continue;
		}
		if (flag) continue;
		if (to != yzh[now][0]){
			node sum; register int u = now;
			for (register int k = real_lgN - 1; k >= 0; --k)
			if (yzh[u][k] && dpt[yzh[u][k]] > dpt[to])
				sum.merge(f[u][k]), u = yzh[u][k];
			if (sum.fi >= dpt[to]){
				vector<pair<int, int> >::iterator it = 
					upper_bound(l.begin(), l.end(), pair<int, int>{dpt[to], inf});
				if (it == l.end() || it -> second >= dpt[to]) continue;
			}
		}
		if (sum[to] > 1 || not_ok.count(dpt[to])) continue;
		if (sum[to] != 0 && (L[now] < L[who[to]] || R[who[to]] < L[now])) continue;
		++ans;
	}
}

signed main(){
	read(n, m), real_lgN = __lg(n) + 2;
	for (int i = 1, u, v; i <= m; ++i) read(u, v), xym.add(u, v), xym.add(v, u);
	dfs(1);
	for (register int k = 1; k < real_lgN; ++k)
	for (register int i = 1; i <= n; ++i){
		yzh[i][k] = yzh[yzh[i][k - 1]][k - 1];
		f[i][k] = f[i][k - 1];
		if (yzh[i][k - 1]) f[i][k].merge(f[yzh[i][k - 1]][k - 1]);
	}
	redfs(1), write(m - ans);
	return 0;
}

TLE code

点击查看代码

//#pragma GCC optimize(3)
//#pragma GCC optimize("Ofast", "inline", "-ffast-math")
//#pragma GCC target("avx", "sse2", "sse3", "sse4", "mmx")
#include <iostream>
#include <cstdio>
#define debug(a) cerr << "Line: " << __LINE__ << " " << #a << endl
#define print(a) cerr << #a << "=" << (a) << endl
#define file(a) freopen(#a".in", "r", stdin), freopen(#a".out", "w", stdout)
#define main Main(); signed main(){ return ios::sync_with_stdio(0), cin.tie(0), Main(); } signed Main
using namespace std;

#include <unordered_set>
#include <algorithm>
#include <vector>
#include <set>

const int inf = 0x3f3f3f3f;
const int  N  = 100010;
const int  M  = 300010;
const int lgN = __lg(N) + 2;

struct Graph{
	struct node{
		int to, nxt;
	} edge[M << 1];
	int eid, head[N];
	inline void add(const int &u, const int &v){
		edge[++eid] = {v, head[u]};
		head[u] = eid;
	}
	inline node & operator [] (const int x){
		return edge[x];
	}
} xym;

struct node{
	int fi, se;
	inline node (int fi = inf, int se = inf): fi(fi), se(se) {}
	inline void merge(const node & o){
		this -> merge(o.fi);
		this -> merge(o.se);
	}
	inline void merge(const int val){
		if (val < fi) se = fi, fi = val;
		else if (fi < val && val < se) se = val;
	}
};

int n, m, ans;
int real_lgN;

int dpt[N], yzh[N][lgN];
int L[N], R[N], timer;
vector<int> son[N], rev[N];
node f[N][lgN], pos[N], s[N];
int sum[N], who[N];

void dfs(int now){
	L[now] = ++timer, dpt[now] = dpt[yzh[now][0]] + 1;
	vector<node> l, r; int cnt = 0;
	for (register int i = xym.head[now]; i; i = xym[i].nxt){
		int to = xym[i].to;
		if (to == yzh[now][0]) continue;
		if (dpt[to]){
			pos[now].merge(dpt[to]);
			if (dpt[to] <= dpt[now]) rev[now].emplace_back(to);
		} else {
			son[now].emplace_back(to), yzh[to][0] = now;
			rev[to].emplace_back(now);
			dfs(to), s[now].merge(s[to]), l.emplace_back(s[to]);
			if (s[to].fi >= dpt[now]) ++sum[now], who[now] = to;
		}
	}
	r = l, cnt = son[now].size();
	for (register int i = cnt - 2; i >= 0; --i) r[i].merge(r[i + 1]);
	for (register int i = 0; i < cnt; ++i){
		register int to = son[now][i];
		(i > 0) && (f[to][0].merge(l[i - 1]), l[i].merge(l[i - 1]), yzh_i_love_you);
		(i + 1 < cnt) && (f[to][0].merge(r[i + 1]), yzh_i_love_you);
		f[to][0].merge(pos[now]);
	}
	R[now] = timer, s[now].merge(pos[now]);
}

int root[N];
struct Combinable_Segment_Tree{
	struct node{
		int lson, rson;
		int minn, maxx;
	} tree[N << 4];
	int tot;
	
	inline node & operator [] (const int x){
		return tree[x];
	}
	
	inline void pushup(int idx){
		if (tree[idx].lson && tree[idx].rson){
			tree[idx].minn = Fast::min(tree[tree[idx].lson].minn, tree[tree[idx].rson].minn);
			tree[idx].maxx = Fast::max(tree[tree[idx].lson].maxx, tree[tree[idx].rson].maxx);
		} else if (tree[idx].lson){
			tree[idx].minn = tree[tree[idx].lson].minn;
			tree[idx].maxx = tree[tree[idx].lson].maxx;
		} else if (tree[idx].rson){
			tree[idx].minn = tree[tree[idx].rson].minn;
			tree[idx].maxx = tree[tree[idx].rson].maxx;
		} else {
			tree[idx].minn = inf, tree[idx].maxx = -inf;
		}
	}
	
	void insert(int &idx, int trl, int trr, int p){
		if (trl > p || trr < p) return;
		if (!idx) tree[idx = ++tot] = {0, 0, inf, -inf};
		if (trl == trr) return tree[idx].minn = tree[idx].maxx = p, void();
		int mid = (trl + trr) >> 1;
		insert(tree[idx].lson, trl, mid, p);
		insert(tree[idx].rson, mid + 1, trr, p);
		pushup(idx);
	}
	
	void erase(int &idx, int trl, int trr, int p){
		if (trl > p || trr < p) return;
		if (!idx) tree[idx = ++tot] = {0, 0, inf, -inf};
		if (trl == trr) return tree[idx].minn = inf, tree[idx].maxx = -inf, void();
		int mid = (trl + trr) >> 1;
		erase(tree[idx].lson, trl, mid, p);
		erase(tree[idx].rson, mid + 1, trr, p);
		pushup(idx);
	}
	
	int combine(int idx, int oidx, int trl, int trr){
		if (!idx || !oidx) return idx | oidx;
		if (trl == trr) return tree[idx].maxx = tree[oidx].maxx, tree[idx].minn = tree[oidx].minn, idx;
		int mid = (trl + trr) >> 1;
		tree[idx].lson = combine(tree[idx].lson, tree[oidx].lson, trl, mid);
		tree[idx].rson = combine(tree[idx].rson, tree[oidx].rson, mid + 1, trr);
		return pushup(idx), idx;
	}
} tree;

void redfs(int now){
	unordered_set<int> not_ok;
	vector<pair<int, int> > l;
	bool flag = false;
	for (const auto& to: son[now]){
		redfs(to);
		while (root[to]){
			int minn = tree[root[to]].minn, maxx = tree[root[to]].maxx;
			if (minn == inf || maxx == -inf) break;
			if (maxx < dpt[now]) break;
			tree.erase(root[to], 1, n, maxx);
		}
		if (root[to]){
			int minn = tree[root[to]].minn, maxx = tree[root[to]].maxx;
			if (minn != inf && maxx != -inf && minn != maxx)
				l.emplace_back(maxx, minn);
		}
		root[now] = tree.combine(root[now], root[to], 1, n);
		if (s[to].fi >= dpt[now]) flag = true;
		else if (s[to].se >= dpt[now]) not_ok.insert(s[to].fi);
	}
	sort(l.begin(), l.end());
	int tot = l.size();
	for (register int i = tot - 2; i >= 0; --i)
		l[i].second = Fast::min(l[i].second, l[i + 1].second);
	for (const auto& to: rev[now]){
		tree.insert(root[now], 1, n, dpt[to]);
		if (to == 1){
			if (sum[to] != 1){
				if (to == yzh[now][0] && son[now].size() == 0u && sum[to] == 2) ++ans;
			} else {
				if (to == yzh[now][0] && son[now].size() != 1u) continue;
				if (to != yzh[now][0] && (flag || not_ok.count(1))) continue;
				++ans;
			}
			continue;
		}
		if (flag) continue;
		if (to != yzh[now][0]){
			node sum; register int u = now;
			for (register int k = real_lgN - 1; k >= 0; --k)
			if (yzh[u][k] && dpt[yzh[u][k]] > dpt[to])
				sum.merge(f[u][k]), u = yzh[u][k];
			if (sum.fi >= dpt[to]){
				vector<pair<int, int> >::iterator it = 
					upper_bound(l.begin(), l.end(), pair<int, int>{dpt[to], inf});
				if (it == l.end() || it -> second >= dpt[to]) continue;
			}
		}
		if (sum[to] > 1 || not_ok.count(dpt[to])) continue;
		if (sum[to] != 0 && (L[now] < L[who[to]] || R[who[to]] < L[now])) continue;
		++ans;
	}
}

signed main(){
	read(n, m), real_lgN = __lg(n) + 2;
	for (int i = 1, u, v; i <= m; ++i) read(u, v), xym.add(u, v), xym.add(v, u);
	dfs(1);
	for (register int k = 1; k < real_lgN; ++k)
	for (register int i = 1; i <= n; ++i){
		yzh[i][k] = yzh[yzh[i][k - 1]][k - 1];
		f[i][k] = f[i][k - 1];
		if (yzh[i][k - 1]) f[i][k].merge(f[yzh[i][k - 1]][k - 1]);
	}
	redfs(1), write(m - ans);
	return 0;
}

posted @ 2024-04-01 18:05 XuYueming 阅读(7) 评论(0) 编辑收藏举报

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[COCI 2023/2024 #1] Mostovi 题解

前言

题目分析

树边

非树边

代码（已略去快读快写）

左偏树

启发式合并

dfs 序加线段树

线段树合并

UPDATE on 2024.4.7

AC code

TLE code

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