无向图上割点/割边

下面的四个算法都是针对无向图的

割点

注意:割点不判断前一个边(判断也行) 割边一定需要判断前一个边

解释:对于 $(u, v)$ 这条边我们如果让 $v$ 更新 $(v, u)$ 这条边这个 $v$ 节点还是只能最靠上回溯到 $u$ 节点对答案是没有影响的

但是割边则不同如果你让 $v$ 更新了 $(v, u)$ 这条边那么这条边的 $l o w [v]$ 将会被错误地更新为 $d f n [u]$ 导致 $l o w [v]$ 永远会大于等于 $d f n [u]$ 导致最后搜不出来割边

$v \neq f a$ 是错的!!!!!!!!!!!!!!必须判断反向边!!!!!!!!!!!!!!

P3388 【模板】割点（割顶）

$t a r j a n$ 基础操作相比于缩点有所不同:

如果一个点 $u$ 为根且有至少两个子树(这两个子树必须是两个独立的子树即互相之间不能到达) 那么是割点
如果一个点 $u$ 不是根且 $l o w [v] \geq d f n [u]$ 说明子节点 $v$ 不能通过非树边到达 $u$ 以上的节点则必然可以通过割掉 $u$ 来使得 $v$ 不能和 $u$ 前面的点连通那么是割点

在 $v$ 节点有 $d f n$ 值的时候我们一定要更新 $l o w [u] = m i n (l o w [u], d f n [v])$ 因为点 $u$ 不一定能到达 $v$ 能到达的所有节点

还需要注意:一个点作为割点可能会被标记多次我们只需要标记在最后统计即可或者可以用 $v e c t o r$ 记录并去重

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define mid (l+r>>1)
#define inl inline
#define eb emplace_back
#define pii pair<int,int>
#define fi first
#define se second
const int N = 2e4 + 5;
const int inf = 0x3f3f3f3f;
char buf[1<<24] , *p1 , *p2;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<24,stdin),p1==p2)?EOF:*p1++)
// #define getchar() cin.get();
int read()
{
	int x = 0 , f = 1;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f ;
}

int n , m , pd[N] , cnt;

vector<int> e[N];
inl void add ( int u , int v ) { e[u].eb(v); }

int dfn[N] , low[N] , timer;

void tarjan ( int u , int rt )
{
	dfn[u] = low[u] = ++timer;
	int child = 0;
	for ( auto v : e[u] )
	{
		if ( !dfn[v] )
		{
			tarjan ( v , rt ) , low[u] = min ( low[u] , low[v] );
			if ( low[v] >= dfn[u] && u != rt ) pd[u] = 1;
			++ child;
		}
		else low[u] = min ( low[u] , dfn[v] );
	}
	if ( u == rt && child >= 2 ) pd[u] = 1;
}

signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr) , cout.tie(nullptr);
	n = read() , m = read();
	for ( int i = 1 , u , v ; i <= m ; i ++ ) u = read() , v = read() , add ( u , v ) , add ( v , u );
	for ( int i = 1 ; i <= n ; i ++ ) if ( !dfn[i] ) tarjan(i,i);
	for ( int i = 1 ; i <= n ; i ++ ) if ( pd[i] ) cnt ++;
	cout << cnt << endl;
	for ( int i = 1 ; i <= n ; i ++ ) if ( pd[i] ) cout << i << ' ';
	return 0;
}

P3469 [POI2008] BLO-Blockade

显然对于每一个割点如果删去它形成的连通块大小分别为 $s_{1 - k}$ 那么答案就是 $\sum_{i = 1}^{k} s_{i} (n - s_{i} - 1)$

(我们对于 $u$ 节点的贡献需要单独计算注意不要忘记 $u$ 没有被删去它本身算一个大小为 $1$ 的连通块)

如果不是割点那么答案即为 $2 * (n - 1)$

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define mid (l+r>>1)
#define inl inline
#define eb emplace_back
#define pii pair<int,int>
#define fi first
#define se second
#define int long long
const int N = 1e5 + 5;
// char buf[1<<24] , *p1 , *p2;
// #define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<24,stdin),p1==p2)?EOF:*p1++)
#define getchar() cin.get();
int read()
{
	int x = 0 , f = 1;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f ;
}

int n , m , ans[N] , cut[N] , sz[N];

vector<int> e[N];
inl void add ( int u , int v ) { e[u].eb(v); }

int dfn[N] , low[N] , timer;

void tarjan ( int u , int rt )
{
	int child = 0 , sum = 0;
	dfn[u] = low[u] = ++ timer , sz[u] = 1;
	for ( auto v : e[u] )
		if ( !dfn[v] )
		{
			tarjan ( v , rt ) , low[u] = min ( low[u] , low[v] ) , sz[u] += sz[v];
			if ( dfn[u] <= low[v] )
			{
				ans[u] += ( n - sz[v] - 1 ) * sz[v] , sum += sz[v];
				if ( u != rt ) cut[u] = 1;
			}
			++ child;
		}
		else low[u] = min ( low[u] , dfn[v] );
	if ( u == rt && child >= 2 ) cut[u] = 1;
	if ( cut[u] ) ans[u] += sum * ( n - sum - 1 );
	ans[u] += 2 * ( n - 1 );
}

signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr) , cout.tie(nullptr);
	n = read() , m = read();
	for ( int i = 1 , u , v ; i <= m ; i ++ ) u = read() , v = read() , add ( u , v ) , add ( v , u );
	for ( int i = 1 ; i <= n ; i ++ ) if ( !dfn[i] ) tarjan ( i , i );
	for ( int i = 1 ; i <= n ; i ++ ) cout << ans[i] << endl;
	return 0;
}

P5058 [ZJOI2004] 嗅探器

为了简化题目我们的 $t a r j a n$ 从其中一个信息中心开始跑即可那么一个合法的割点当且仅当 $l o w [v] \geq d f n [u]$ 且 $d f n [v v] \geq d f n [v]$

这里第二个柿子是因为我们现在看的是割掉 $u$ 点能否让 $(u, v)$ 这条边能连通的子树内的所有点变得不联通了

那么 $d f n [v v] \geq d f n [v]$ 当且仅当 $v v$ 在 $u$ 子树中(不包括 $u$ )

这样求出所有割点即可

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define mid (l+r>>1)
#define inl inline
#define eb emplace_back
#define pii pair<int,int>
#define fi first
#define se second
const int N = 2e5 + 5;
//char buf[1<<24] , *p1 , *p2;
//#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<24,stdin),p1==p2)?EOF:*p1++)
 #define getchar() cin.get();
int read()
{
	int x = 0 , f = 1;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f ;
}

int n , uu , vv , cut[N];

vector<int> e[N];
inl void add ( int u , int v ) { e[u].eb(v); }

int dfn[N] , low[N] , timer;

void tarjan ( int u , int rt )
{
	dfn[u] = low[u] = ++timer; 
	for ( auto v : e[u] )
	{
		if ( !dfn[v] )
		{
			tarjan(v,rt) , low[u] = min ( low[u] , low[v] );
			if ( low[v] >= dfn[u] && u != rt && dfn[vv] >= dfn[v] ) cut[u] = 1;
		}
		else low[u] = min ( low[u] , dfn[v] );		
	}
}

signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr) , cout.tie(nullptr);
	n = read();
	while ( ( uu = read() ) + ( vv = read() ) != 0 ) add ( uu , vv ) , add ( vv , uu );
	uu = read() , vv = read();
	tarjan(uu,uu);
	for ( int i = 1 ; i <= n ; i ++ ) if ( cut[i] ) { cout << i << endl; return 0; }
	return cout << "No solution" << endl , 0;
}

割边

相当于在割点板子上改了一下判断条件 $l o w [v] > d f n [u]$ 而且需要特判反向边

这样是为了防止搜索 $v$ 的时候将 $l o w [v]$ 错误地更新为了 $d f n [u]$ 导致 $l o w [v]$ 永远会大于等于 $d f n [u]$ 导致最后搜不出来割边

P1656 炸铁路

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define mid (l+r>>1)
#define inl inline
#define eb emplace_back
#define pii pair<int,int>
#define fi first
#define se second
const int N = 1e5 + 5;
char buf[1<<24] , *p1 , *p2;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<24,stdin),p1==p2)?EOF:*p1++)
// #define getchar() cin.get();
int read()
{
	int x = 0 , f = 1;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f ;
}

int n , m , k , l , sza[N] , szb[N] , cnt = 1;

vector<pii> e[N] , ans;
inl void add ( int u , int v , int w ) { e[u].eb(v,w); }

int low[N] , dfn[N] , timer;

void tarjan ( int u , int ff )
{
	dfn[u] = low[u] = ++timer;
	for ( auto [v,p] : e[u] )
	{
		if ( !dfn[v] )
		{
			tarjan ( v , p ) , low[u] = min ( low[u] , low[v] ) , sza[u] += sza[v] , szb[u] += szb[v];
			if ( low[v] > dfn[u] && ( sza[v] == 0 || szb[v] == 0 || sza[v] == k || szb[v] == l ) ) ans.eb(u,v);
		}
		else if ( p ^ ( ff ^ 1 ) ) low[u] = min ( low[u] , dfn[v] );
	}
}

signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr) , cout.tie(nullptr);
	n = read() , m = read() , k = read() , l = read();
	for ( int i = 1 ; i <= k ; i ++ ) sza[read()] = 1;
	for ( int i = 1 ; i <= l ; i ++ ) szb[read()] = 1;
	for ( int i = 1 , u , v ; i <= m ; i ++ ) u = read() , v = read() , add ( u , v , ++ cnt ) , add ( v , u , ++ cnt );
	tarjan(1,0);
	cout << ans.size() << endl;
	for ( auto p : ans ) cout << p.fi << ' ' << p.se << endl;
	return 0;
}

P7687 [CEOI2005] Critical Network Lines

割边板子

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define mid (l+r>>1)
#define inl inline
#define eb emplace_back
#define pii pair<int,int>
#define fi first
#define se second
const int N = 1e5 + 5;
char buf[1<<24] , *p1 , *p2;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<24,stdin),p1==p2)?EOF:*p1++)
// #define getchar() cin.get();
int read()
{
	int x = 0 , f = 1;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f ;
}

int n , m , k , l , sza[N] , szb[N] , cnt = 1;

vector<pii> e[N] , ans;
inl void add ( int u , int v , int w ) { e[u].eb(v,w); }

int low[N] , dfn[N] , timer;

void tarjan ( int u , int ff )
{
	dfn[u] = low[u] = ++timer;
	for ( auto [v,p] : e[u] )
	{
		if ( !dfn[v] )
		{
			tarjan ( v , p ) , low[u] = min ( low[u] , low[v] ) , sza[u] += sza[v] , szb[u] += szb[v];
			if ( low[v] > dfn[u] && ( sza[v] == 0 || szb[v] == 0 || sza[v] == k || szb[v] == l ) ) ans.eb(u,v);
		}
		else if ( p ^ ( ff ^ 1 ) ) low[u] = min ( low[u] , dfn[v] );
	}
}

signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr) , cout.tie(nullptr);
	n = read() , m = read() , k = read() , l = read();
	for ( int i = 1 ; i <= k ; i ++ ) sza[read()] = 1;
	for ( int i = 1 ; i <= l ; i ++ ) szb[read()] = 1;
	for ( int i = 1 , u , v ; i <= m ; i ++ ) u = read() , v = read() , add ( u , v , ++ cnt ) , add ( v , u , ++ cnt );
	tarjan(1,0);
	cout << ans.size() << endl;
	for ( auto p : ans ) cout << p.fi << ' ' << p.se << endl;
	return 0;
}

点双

点双/边双针对无向图

在 $t a r j a n$ 中如果我们搜到了一个割点那么将现在栈中的所有节点全部弹出来但是根节点不能出栈因为割点一定属于多个点双

P8435 【模板】点双连通分量

代码中的特判是针对孤立点的() 如果我们想要用有没有边来判断孤立点的话记得需要去重边

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define mid (l+r>>1)
#define inl inline
#define eb emplace_back
#define ls p<<1
#define rs p<<1|1
#define lson ls,l,mid
#define rson rs,mid+1,r
#define getchar() cin.get();
const int N = 5e5 + 5;
int read()
{
	int x = 0 , f = 1;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f ;
}

int n , m , mod , a[N] , tot;

vector<int> e[N] , bcc[N];
inl void add ( int u , int v ) { e[u].eb(v); }

int low[N] , dfn[N] , timer;
int sta[N] , top;

void tarjan ( int u , int rt )
{
	low[u] = dfn[u] = ++timer;
	sta[++top] = u;
	for ( auto v : e[u] )
		if ( !dfn[v] ) 	
		{
			tarjan ( v , rt ) , low[u] = min ( low[u] , low[v] );
			if ( dfn[u] <= low[v] )
			{
				++ tot;
				while ( top )
				{
					int x = sta[top--]; bcc[tot].eb(x);
					if ( x == v ) break;
				} 
				bcc[tot].eb(u);
			}
		}
		else low[u] = min ( low[u] , dfn[v] ); 
	if ( u == rt && !e[u].size() ) bcc[++tot].eb(u);
}

signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr) , cout.tie(nullptr);
	n = read() , m = read();
	for ( int i = 1 , u , v ; i <= m ; i ++ )
	{
		u = read() , v = read();
		if ( u != v ) add ( u , v ) , add ( v , u );
	}
	for ( int i = 1 ; i <= n ; i ++ ) if ( !dfn[i] ) tarjan ( i , i );
	cout << tot << endl;
	for ( int i = 1 ; i <= tot ; i ++ )
	{
		cout << bcc[i].size() << ' ';
		for ( auto v : bcc[i] ) cout << v << ' ';
		cout << endl;
	}
	return 0;
}

边双

相当于是将题中所有的割边都去掉形成的联通块那么开一个栈来进行类 $t a r j a n$ 统计即可

而且我们可能加了两条及以上 $(u, v)$ 边这是可以满足从 $u$ 通过返祖边回到 $f a t h e r$ 的那么如果我们只判断 $f a t h e r$ 就扼杀了另一条合法 $(u, v)$ 的可能

所以我们需要记录进边的值来判断是否向下搜索

P8436 【模板】边双连通分量

#include<bits/stdc++.h>
using namespace std;
#define mid ((l+r)>>1)
#define inl inline
#define eb emplace_back
#define endl '\n'
#define ls p<<1
#define rs p<<1|1
#define lson ls,l,mid
#define rson rs,mid+1,r
#define pii pair<int,int>
#define fi first
#define se second
#define mkp make_pair
#define print(x) cerr<<#x<<'='<<x<<endl
#define getchar() cin.get()
const int N = 5e5 + 5;
int read()
{
	int x = 0 , f = 1;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f;
}

int n , m , sta[10000000] , top , cnt = 1;

vector<pii> e[N];
vector<int> bcc[N];
inl void add ( int u , int v , int w ) { e[u].eb(v,w); }

int tot , dfn[N] , low[N] , timer;

void tarjan ( int u , int ff )
{
	low[u] = dfn[u] = ++timer;
	sta[++top] = u;
	for ( auto [v,k] : e[u] )
	{
		if ( !dfn[v] ) tarjan ( v , k ) , low[u] = min ( low[u] , low[v] );
		else if ( k != ( ff ^ 1 ) ) low[u] = min ( low[u] , dfn[v] );
	}
	if ( low[u] == dfn[u] )
	{
		++ tot;
		while ( top )
		{
			int x = sta[top--]; bcc[tot].eb(x);
			if ( x == u ) break;
		}
	}
}

signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(0) , cout.tie(0);
	n = read() , m = read();
	for ( int i = 1 , u , v ; i <= m ; i ++ ) u = read() , v = read() , add ( u , v , ++ cnt ) , add ( v , u , ++ cnt );
	for ( int i = 1 ; i <= n ; i ++ ) if ( !dfn[i] ) tarjan ( i , 0 );
	cout << tot << endl;
	for ( int i = 1 ; i <= tot ; i ++ )
	{
		cout << bcc[i].size() << ' ';
		for ( auto v : bcc[i] )
			cout << v << ' ' ;
		cout << endl;
	}
	return 0;
}

P2860 [USACO06JAN] Redundant Paths G

缩点之后统计一下入度大小

如果叶子节点 $s$ 为偶数那么两两连边即可答案是 $s / 2$

如果是奇数那么是 $s / 2 + 1$

#include<bits/stdc++.h>
using namespace std;
#define mid ((l+r)>>1)
#define inl inline
#define eb emplace_back
#define endl '\n'
#define ls p<<1
#define rs p<<1|1
#define lson ls,l,mid
#define rson rs,mid+1,r
#define pii pair<int,int>
#define fi first
#define se second
#define mkp make_pair
#define print(x) cerr<<#x<<'='<<x<<endl
#define getchar() cin.get()
const int N = 1e5 + 5;
int read()
{
	int x = 0 , f = 1;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f;
}

int n , m , cnt = 1 , u[N] , v[N] , ans , dag[N];

int dfn[N] , low[N] , timer;
int sta[N] , top;
int id[N] , tot;

vector<pii> e[N];
inl void add ( int u , int v , int w ) { e[u].eb(v,w); }

void tarjan ( int u , int ff )
{
	low[u] = dfn[u] = ++timer;
	sta[++top] = u;
	for ( auto [v,k] : e[u] )
	{
		if ( !dfn[v] ) tarjan ( v , k ) , low[u] = min ( low[u] , low[v] );
		else if ( k ^ ( ff ^ 1 ) ) low[u] = min ( low[u] , dfn[v] );
	}
	if ( low[u] == dfn[u] )
	{
		++ tot;
		while ( top )
		{
			int x = sta[top--];
			id[x] = tot;
			if ( x == u ) break;
		}
	}
}


signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(0) , cout.tie(0);
	n = read() , m = read();
	for ( int i = 1 ; i <= m ; i ++ ) u[i] = read() , v[i] = read() , add ( u[i] , v[i] , ++ cnt ) , add ( v[i] , u[i] , ++ cnt ); 
	for ( int i = 1 ; i <= n ; i ++ ) if ( !dfn[i] ) tarjan ( i , 0 );
	for ( int i = 1 ; i <= m ; i ++ ) if ( id[u[i]] != id[v[i]] ) ++ dag[id[u[i]]] , ++ dag[id[v[i]]];
	for ( int i = 1 ; i <= tot ; i ++ ) ans += ( dag[i] == 1 );
	return cout << ans / 2 + ( ans & 1 ) << endl , 0;
}

圆方树

广义圆方树即为:在无向图中将所有点双缩起来作为一个方点放在树中圆点就是树上原有的节点

P4320 道路相遇

题意显然可以转化为 $q$ 次询问 $(u, v)$ 之间的割点数量

那么建立圆方树并树上差分统计即可统计的是圆点的数量

#include<bits/stdc++.h>
using namespace std;
#define mid ((l+r)>>1)
#define inl inline
#define eb emplace_back
#define endl '\n'
#define ls p<<1
#define rs p<<1|1
#define lson ls,l,mid
#define rson rs,mid+1,r
#define pii pair<int,int>
#define fi first
#define se second
#define mkp make_pair
#define print(x) cerr<<#x<<'='<<x<<endl
#define getchar() cin.get()
const int N = 1e6 + 5;
int read()
{
	int x = 0 , f = 1;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f;
}

int n , m , q , cnt , block , w[N] , sum[N];

int dfn[N] , low[N] , timer;
int sta[N] , tp;

vector<int> e[N] , ee[N];
inl void add ( int u , int v ) { e[u].eb(v); }
inl void adde ( int u , int v ) { ee[u].eb(v); }

void tarjan ( int u )
{
	dfn[u] = low[u] = ++timer;
	sta[++tp] = u;
	for ( auto v : e[u] )
	{
		if ( !dfn[v] )
		{
			tarjan ( v ) , low[u] = min ( low[u] , low[v] );
			if ( low[v] >= dfn[u] )
			{
				++ block;
				while ( tp )
				{	
					int x = sta[tp--];
					adde ( block , x ) , adde ( x , block ) , w[x] = 1;
					if ( x == v ) break;
				}
				adde ( block , u ) , adde ( u , block ) , w[u] = 1;
			}
		}		
		else low[u] = min ( low[u] , dfn[v] );
	}
}

int fa[N] , dep[N] , sz[N] , son[N];
int top[N] , rev[N] , pos[N];

struct LCA
{
	void dfs1 ( int u , int ff )
	{
		sum[u] = sum[fa[u]=ff] + w[u] , dep[u] = dep[ff] + 1 , sz[u] = 1;
		for ( auto v : ee[u] )
			if ( v ^ ff )
			{
				dfs1 ( v , u );
				sz[u] += sz[v];
				if ( sz[son[u]] < sz[v] ) son[u] = v;
			}
	}
	void dfs2 ( int u , int tp )
	{
		top[u] = tp , pos[u] = ++timer , rev[timer] = u;
		if ( son[u] ) dfs2 ( son[u] , tp );
		for ( auto v : ee[u] ) if ( v ^ fa[u] && v ^ son[u] ) dfs2 ( v , v );
	}
	int lca ( int u , int v )
	{
		while ( top[u] != top[v] )
		{
			if ( dep[top[u]] < dep[top[v]] ) swap ( u , v );
			u = fa[top[u]];
		}
		if ( dep[u] < dep[v] ) swap ( u , v );
		return v;
	}
}L;

int query ( int u , int v )
{
	int lcaa = L.lca(u,v);
	return sum[u] + sum[v] - sum[lcaa] - sum[fa[lcaa]];
}


signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(0) , cout.tie(0);
	n = read() , m = read() , block = n;
	for ( int i = 1 , u , v ; i <= m ; i ++ ) u = read() , v = read() , add ( u , v ) , add ( v , u );
	for ( int i = 1 ; i <= n ; i ++ ) if ( !dfn[i] ) tarjan ( i );
	timer = 0;
	L.dfs1 ( 1 , 0 ) , L.dfs2 ( 1 , 1 );
	q = read();
	for ( int i = 1 ; i <= q ; i ++ )
	{
		int u = read() , v = read();
		cout << query ( u , v ) << endl;
	}
	return 0;
}

P4606 [SDOI2018] 战略游戏

显然我们求的是一个集合中任意两点的路径并的子图中的白点数量除 $2$ 再减去点集的大小(不重复)

如果直接求点的并的话不好统计

那么我们将节点的权值放到它和它父亲连的边上那么所有路径上不重复白点的个数就是每两个 $d f s$ 序相邻的点的路径的并除 $2$ 再减去点集的大小 (画图可知)

但是这样 $1$ 和 $n$ 节点的 $l c a$ 不会被统计到特判一下即可

注意我们在新树中排序的时候注意要用新树的 $p o s$ 数组来排序而不是老树的 $d f n$ 数组

#include<bits/stdc++.h>
using namespace std;
#define mid ((l+r)>>1)
#define inl inline
#define eb emplace_back
#define endl '\n'
#define ls p<<1
#define rs p<<1|1
#define lson ls,l,mid
#define rson rs,mid+1,r
#define pii pair<int,int>
#define fi first
#define se second
#define mkp make_pair
#define print(x) cerr<<#x<<'='<<x<<endl
#define getchar() cin.get()
const int N = 1e6 + 5;
int read()
{
	int x = 0 , f = 1;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f;
}

int n , m , q , cnt , block , w[N] , ans , a[N];

int dfn[N] , low[N] , timer;
int sta[N] , tp;

vector<int> e[N] , ee[N];
inl void add ( int u , int v ) { e[u].eb(v); }
inl void adde ( int u , int v ) { ee[u].eb(v); }

void tarjan ( int u )
{
	dfn[u] = low[u] = ++timer;
	sta[++tp] = u;
	for ( auto v : e[u] )
		if ( !dfn[v] )
		{
			tarjan(v) , low[u] = min ( low[u] , low[v] );
			if ( low[v] >= dfn[u] )
			{
				++ block;
				while ( tp )
				{
					int x = sta[tp--];
					adde ( block , x ) , adde ( x , block ) , w[x] = 1;
					if ( x == v ) break;
				}
				adde ( block , u ) , adde ( u , block ) , w[u] = 1;
			}
		}
		else low[u] = min ( low[u] , dfn[v] );
}

int sz[N] , fa[N] , dep[N] , sum[N] , son[N];
int tim , rev[N] , pos[N] , top[N];

struct node 
{
	void dfs1 ( int u , int ff )
	{
		fa[u] = ff , sum[u] = sum[ff] + w[u] , dep[u] = dep[ff] + 1 , sz[u] = 1;
		for ( auto v : ee[u] )
			if ( v ^ ff )
			{
				dfs1 ( v , u );
				sz[u] += sz[v];
				if ( sz[son[u]] < sz[v] ) son[u] = v;
			}
	}
	void dfs2 ( int u , int tp )
	{
		top[u] = tp , pos[u] = ++ tim , rev[tim] = u;
		if ( son[u] ) dfs2 ( son[u] , tp );
		for ( auto v : ee[u] ) if ( v ^ fa[u] && v ^ son[u] ) dfs2 ( v , v );
	}
	int lca ( int u , int v )
	{
		while ( top[u] != top[v] )
		{
			if ( dep[top[u]] < dep[top[v]] ) swap ( u , v );
			u = fa[top[u]];
		}
		if ( dep[u] < dep[v] ) swap ( u , v );
		return v;
	}
}L;

void init()
{
	for ( int i = 1 ; i <= 2 * n ; i ++ ) e[i].clear();
	for ( int i = 1 ; i <= 2 * n ; i ++ ) ee[i].clear();
	memset ( son , 0 , sizeof son );
	memset ( w , 0 , sizeof w );
	memset ( dfn , 0 , sizeof dfn );
	memset ( sum , 0 , sizeof sum );
	tim = timer = tp = ans = 0;
}

int dis ( int u , int v )
{
	int lcaa = L.lca ( u , v );
	return sum[u] +	sum[v] - 2 * sum[lcaa];
}

signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(0) , cout.tie(0);
	int T = read();
	while ( T -- )
	{
		init();
		n = read() , m = read() , block = n;
		for ( int i = 1 , u , v ; i <= m ; i ++ ) u = read() , v = read() , add ( u , v ) , add ( v , u );
		for ( int i = 1 ; i <= n ; i ++ ) if ( !dfn[i] ) tarjan(i);
		L.dfs1 ( 1 , 0 ) , L.dfs2 ( 1 , 1 );
		q = read();	
		for ( int i = 1 ; i <= q ; i ++ )
		{
			ans = 0;
			int s = read();
			for ( int j = 1 ; j <= s ; j ++ ) a[j] = read();
			sort ( a + 1 , a + s + 1 , [](const int &a , const int &b) { return pos[a] < pos[b]; } );
			a[s+1] = a[1];
			for ( int j = 1 ; j <= s ; j ++ ) ans += dis ( a[j] , a[j+1] );
			ans >>= 1 , ans -= s;
			ans += w[L.lca(a[1],a[s])];
			cout << ans << endl;
		}
	}
	return 0;
}

P3854 [TJOI2008] 通讯网破坏

相当于在圆方树上询问是否第三个点是否在第一个点和第二个点的路径上类比仓鼠那道题即可得解

#include<bits/stdc++.h>
using namespace std;
#define mid ((l+r)>>1)
#define inl inline
#define eb emplace_back
#define endl '\n'
#define ls p<<1
#define rs p<<1|1
#define lson ls,l,mid
#define rson rs,mid+1,r
#define pii pair<int,int>
#define fi first
#define se second
#define mkp make_pair
#define print(x) cerr<<#x<<'='<<x<<endl
#define getchar() cin.get()
const int N = 1e6 + 5;
int read()
{
	int x = 0 , f = 1;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f;
}

int n , m , q , cnt , block , w[N] , ans , a[N];

int dfn[N] , low[N] , timer;
int sta[N] , tp;

vector<int> e[N] , ee[N];
inl void add ( int u , int v ) { e[u].eb(v); }
inl void adde ( int u , int v ) { ee[u].eb(v); }

void tarjan ( int u )
{
	dfn[u] = low[u] = ++timer;
	sta[++tp] = u;
	for ( auto v : e[u] )
		if ( !dfn[v] )
		{
			tarjan(v) , low[u] = min ( low[u] , low[v] );
			if ( low[v] >= dfn[u] )
			{
				++ block;
				while ( tp )
				{
					int x = sta[tp--];
					adde ( block , x ) , adde ( x , block ) , w[x] = 1;
					if ( x == v ) break;
				}
				adde ( block , u ) , adde ( u , block ) , w[u] = 1;
			}
		}
		else low[u] = min ( low[u] , dfn[v] );
}

int sz[N] , fa[N] , dep[N] , sum[N] , son[N];
int tim , rev[N] , pos[N] , top[N];

struct node 
{
	void dfs1 ( int u , int ff )
	{
		fa[u] = ff , sum[u] = sum[ff] + w[u] , dep[u] = dep[ff] + 1 , sz[u] = 1;
		for ( auto v : ee[u] )
			if ( v ^ ff )
			{
				dfs1 ( v , u );
				sz[u] += sz[v];
				if ( sz[son[u]] < sz[v] ) son[u] = v;
			}
	}
	void dfs2 ( int u , int tp )
	{
		top[u] = tp , pos[u] = ++ tim , rev[tim] = u;
		if ( son[u] ) dfs2 ( son[u] , tp );
		for ( auto v : ee[u] ) if ( v ^ fa[u] && v ^ son[u] ) dfs2 ( v , v );
	}
	int lca ( int u , int v )
	{
		while ( top[u] != top[v] )
		{
			if ( dep[top[u]] < dep[top[v]] ) swap ( u , v );
			u = fa[top[u]];
		}
		if ( dep[u] < dep[v] ) swap ( u , v );
		return v;
	}
}L;

int dis ( int u , int v )
{
	int lcaa = L.lca ( u , v );
	return dep[u] +	dep[v] - 2 * dep[lcaa];
}

signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(0) , cout.tie(0);
	n = read() , m = read() , block = n;
	for ( int i = 1 , u , v ; i <= m ; i ++ ) u = read() , v = read() , add ( u , v ) , add ( v , u );
	for ( int i = 1 ; i <= n ; i ++ ) if ( !dfn[i] ) tarjan(i);
	L.dfs1 ( 1 , 0 ) , L.dfs2 ( 1 , 1 );
	q = read();	
	for ( int i = 1 ; i <= q ; i ++ )
	{
		int s = read() , t = read() , c = read();
		if ( dis ( s , t ) == dis ( s , c ) + dis ( c , t ) ) puts("yes");
		else puts("no");
	}
	return 0;
}

posted @ 2023-08-31 20:44 Echo_Long 阅读(28) 评论(0) 编辑收藏举报

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Echo-Long

无向图上割点/割边

无向图上割点/割边

割点

P3388 【模板】割点（割顶）

P3469 [POI2008] BLO-Blockade

P5058 [ZJOI2004] 嗅探器

割边

P1656 炸铁路

P7687 [CEOI2005] Critical Network Lines

点双

P8435 【模板】点双连通分量

边双

P8436 【模板】边双连通分量

P2860 [USACO06JAN] Redundant Paths G

圆方树

P4320 道路相遇

P4606 [SDOI2018] 战略游戏

P3854 [TJOI2008] 通讯网破坏

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