[学习笔记] 强连通分量

Find Problems：

一、DFS Forest

从这张经典图说起：

在给定的有向图 \(G = (V, E)\) 内，遍历这张图，其中边分为 \(4\) 种：

树边（tree edge）：黑色的边，每一次从 \(u\) 访问到一个未访问的节点 \(v\)，则称 \((u, v)\) 为树边。
返祖边（back edge）：红色的边，每一次从 \(u\) 回溯到一个 \(u\) 的已经访问的祖先 \(v\)，则称 \((u, v)\) 为 返祖边。
横叉边（cross edge）：蓝色的边，每一次 \(u\) 访问到一个已经访问过的节点 \(v\)，但 \(v\) 不是 \(u\) 的祖先，则称 \((u, v)\) 为 横叉边。
前向边（forward edge）：绿色的边，每一次 \(u\) 访问到一个还没有访问过的节点 \(v\)，并且 \(u\) 是 \(v\) 的祖先，则称 \((u, v)\) 为 前向边。

其中第一个被访问的节点 \(rt\) 被称为 DFS Forest 的根。

二、强连通分量

可达是指 \(u, v\) 之间至少有一条路径。
弱连通（weakly connected） 是指 \(u, v\) 两点，\(u, v\) 可达或 \(v, u\) 可达至少满足一个。
弱连通图（weakly connected graph） 是指将一张有向图的有向边换成无向边，可以构成一个连通块的有向图。
强连通（strongly connected） 是指 \(u, v\) 两点，\(u, v\) 可达并且 \(v, u\) 可达。
强连通图（strongly connected graph） 是指有向图中任意两点 \(u, v(u \neq v)\) 都是强连通的。
强连通分量（strongly connected component） 是指有向图中极大强连通图。

性质：若 \(u, v\) 强连通，且 \(v, w\) 强连通，则 \(u, w\) 强连通（具有传递性）。

三、Tarjan

性质 1：强连通分量的分布一定是有向图切掉一些边，形成一堆连通块，这些连通块就是强连通分量。
性质 2：DFS Forest 上如果一个连通分量没有边连出，那么这个连通分量就是一个 强连通分量。
\(stk\)：一个栈，还没有被切掉的节点。
\(dfn_u\)：节点 \(u\) 的 DFS 序。
\(low_u\)：在 \(u\) 的子树内能回溯到的最早的在栈中的节点。

然后从 \(u\) 到 \(v\) dfs 有三种情况：

\(v\) 未被访问：继续 dfs，low[u] = min (low[u], low[v])。
\(v\) 已被访问，并且在 \(stk\) 中，low[u] = min (low[u], dfn[v])。
\(v\) 已被访问，不在 \(stk\) 中，就当无事发生。

如果 \(dfn_u = low_u\)，容易发现 \(u\) 已经寄了，所以栈中 \(u\) 以上的（含 \(u\)）的所有节点都构成一个强连通分量。

求 SCC 的板子（洛谷 B3609）代码：

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/hash_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
typedef long long ll;
typedef pair < int, int > PII;
typedef int itn;
mt19937 RND_MAKER (chrono :: steady_clock :: now ().time_since_epoch ().count ());
inline ll randomly (const ll l, const ll r) {return (RND_MAKER () ^ (1ull << 63)) % (r - l + 1) + l;}
bool Memory_Begins;
const double pi = acos (-1);
//__gnu_pbds :: tree < Key, Mapped, Cmp_Fn = std :: less < Key >, Tag = rb_tree_tag, Node_Upadte = null_tree_node_update, Allocator = std :: allocator < char > > ;
//__gnu_pbds :: tree < PPS, __gnu_pbds :: null_type, less < PPS >, __gnu_pbds :: rb_tree_tag, __gnu_pbds :: tree_order_statistics_node_update > tr;
template < class Z >
inline void read (Z &tmp) {
	Z x = 0, f = 0;
	char c = getchar ();
	for ( ; c < '0' || c > '9' ; c = getchar ()) f |= (c == '-');
	for ( ; c >= '0' && c <= '9' ; c = getchar ()) x = (x << 1) + (x << 3) + (c & 15);
	tmp = !f ? x : -x;
}
const int N = 1e5 + 5;
vector < int > G[N];
int n, m, dfn[N], low[N], isin[N], time_click, scc[N], cnt;
vector < int > sccs[N];
stack < int > stk;
inline void tarjan (int u) {
	low[u] = dfn[u] = ++ time_click;
	isin[u] = 1;
	stk.push (u);
	for (int v : G[u]) {
		if (!dfn[v]) {tarjan (v), low[u] = min (low[u], low[v]);}
		else {if (isin[v]) low[u] = min (low[u], dfn[v]);}
	}
	if (dfn[u] == low[u]) {
		cnt ++;
		while (true) {
			int v = stk.top (); scc[v] = cnt, isin[v] = 0, stk.pop ();
			if (u == v) break;
		}
	}
}
bool Memory_Ends;
signed main () {
	fprintf (stderr, "%.3lf MB\n", (&Memory_Begins - &Memory_Ends) / 1048576.0);
	read (n), read (m);
	for (int i = 1;i <= m; ++ i) {
		int u, v;
		read (u), read (v);
		G[u].push_back (v);
	} 
	for (int i = 1;i <= n; ++ i) {if (!dfn[i]) tarjan (i);}
	for (int i = 1;i <= n; ++ i) sccs[scc[i]].push_back (i);
	set < int > st; st.clear ();
	printf ("%d\n", cnt);
	for (int i = 1;i <= n; ++ i) {
		if (st.count (i)) continue;
		for (int u : sccs[scc[i]]) printf ("%d ", u), st.insert (u);
		printf ("\n");
	}
	fprintf (stderr, "%.3lf ms\n", 1e3 * clock () / CLOCKS_PER_SEC);
	return 0;
}
/*
Things to check:
1. int ? long long ? unsigned int ? unsigned long long ?
2. array size ? is it enough ?
*/

四、Codeforces 427C - Checkposts

http://codeforces.com/problemset/problem/427/C

首先我们求出强连通分量，设有 \(k\) 个，分别为 \(\{u_{1, 1}, u_{1, 2}, \dots, u_{1, |u_1|}\}\)，\(\{u_{2, 1}, u_{2, 2}, \dots, u_{2, |u_2|}\}\)，...，\(\{u_{k, 1}, u_{k, 2}, \dots, u_{k, |u_k|}\}\)。

贪心，可得每一个强连通分量内，建造的花费为 SCC 内部的点的最小花费，i. e.

设 \(w_i\) 为保证 \(i\) 的安全的花费，那么最小花费为：

\[\sum_{i = 1}^{k} \min_{j = 1}^{|u_i|}w_{u_{i,j}} \]

肯定，每个连通块要取最小的，那么根据乘法原理，肯定是每个 SCC 中 \(w\) 为 \(\min w\) 的数量，即：

\[\prod_{i = 1}^{k} \sum_{j = 1}^{|u_i|} \left[w_{u_{i,j}} = \min_{x=1}^{|u_i|}w_{u_{i,x}}\right] \]

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/hash_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
typedef long long ll;
typedef pair < int, int > PII;
typedef int itn;
mt19937 RND_MAKER (chrono :: steady_clock :: now ().time_since_epoch ().count ());
inline ll randomly (const ll l, const ll r) {return (RND_MAKER () ^ (1ull << 63)) % (r - l + 1) + l;}
bool Memory_Begins;
const double pi = acos (-1);
//__gnu_pbds :: tree < Key, Mapped, Cmp_Fn = std :: less < Key >, Tag = rb_tree_tag, Node_Upadte = null_tree_node_update, Allocator = std :: allocator < char > > ;
//__gnu_pbds :: tree < PPS, __gnu_pbds :: null_type, less < PPS >, __gnu_pbds :: rb_tree_tag, __gnu_pbds :: tree_order_statistics_node_update > tr;
template < class Z >
inline void read (Z &tmp) {
	Z x = 0, f = 0;
	char c = getchar ();
	for ( ; c < '0' || c > '9' ; c = getchar ()) f |= (c == '-');
	for ( ; c >= '0' && c <= '9' ; c = getchar ()) x = (x << 1) + (x << 3) + (c & 15);
	tmp = !f ? x : -x;
}
const int N = 1e5 + 5;
vector < int > G[N];
int n, m, dfn[N], low[N], isin[N], time_click, scc[N], cnt;
vector < int > st[N];
stack < int > stk;
inline void tarjan (int u) {
	low[u] = dfn[u] = ++ time_click;
	isin[u] = 1;
	stk.push (u);
	for (int v : G[u]) {
		if (!dfn[v]) {tarjan (v), low[u] = min (low[u], low[v]);}
		else {if (isin[v]) low[u] = min (low[u], dfn[v]);}
	}
	if (dfn[u] == low[u]) {
		cnt ++;
		while (true) {
			int v = stk.top (); scc[v] = cnt, isin[v] = 0, stk.pop ();
			if (u == v) break;
		}
	}
}
ll w[N], val[N];
const ll mod = 1e9 + 7;
bool Memory_Ends;
signed main () {
	fprintf (stderr, "%.3lf MB\n", (&Memory_Begins - &Memory_Ends) / 1048576.0);
	read (n);
	for (int i = 1;i <= n; ++ i) read (w[i]);
	read (m);
	for (int i = 1;i <= m; ++ i) {
		int u, v;
		read (u), read (v);
		G[u].push_back (v);
	} 
	for (int i = 1;i <= n; ++ i) {if (!dfn[i]) tarjan (i);}
	for (int i = 1;i <= cnt; ++ i) val[i] = 2e9;
	for (int i = 1;i <= n; ++ i) val[scc[i]] = min (val[scc[i]], w[i]);
	ll ans1 = 0, ans2 = 1;
	for (int i = 1;i <= cnt; ++ i) ans1 += val[i];
	for (int i = 1;i <= n; ++ i) st[scc[i]].push_back (i);
	for (int i = 1;i <= cnt; ++ i) {
		ll jd = 0;
		for (int u : st[i]) {
			if (w[u] == val[i]) jd ++;
		}
		ans2 = ans2 * jd % mod;
	}
	printf ("%lld %lld\n", ans1, ans2);
	fprintf (stderr, "%.3lf ms\n", 1e3 * clock () / CLOCKS_PER_SEC);
	return 0;
}
/*
Things to check:
1. int ? long long ? unsigned int ? unsigned long long ?
2. array size ? is it enough ?
*/

五、洛谷 P2272 [ZJOI2007] 最大半连通子图

https://www.luogu.com.cn/problem/P2272

首先先考虑一个弱化版本：原图是 DAG。

那么这个导出子图就应该是一条链，如果有分叉就是这个形状：

这样就会导致 \(2, 3\) 相互不可达。

所以答案就分别是：最长链长度，最长链数量。

然后就记 \(f_u\) 为终点为 \(u\) 的最长链长度，\(g_u\) 是终点为 \(u\) 的最长链数量。

那么从 \(u\) 转移到 \(v\) 的转移方程就是：

\[f_v = \max(f_v, f_u + 1) \]

\[g_v = \begin{cases}g_v, f_v > f_u + 1\\g_u + g_v, f_v = f_u + 1\\ g_u, f_v < f_u + 1\end{cases} \]

然后就是个 \(f\) 的最大值计数（权值为 \(g\) 的值）。

回到原问题：很简单，直接缩点，就是 DAG，只不过每个点的贡献就不是 \(1\) 了，而是缩点后每个点代表的原来的点的数量。

缩点后一定要判重边 & 自环！！1

还有就是所有 \(v\) 的入边都转移完了再去转移 \(v\) 的出边。

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/hash_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
typedef long long ll;
typedef pair < int, int > PII;
typedef int itn;
mt19937 RND_MAKER (chrono :: steady_clock :: now ().time_since_epoch ().count ());
inline ll randomly (const ll l, const ll r) {return (RND_MAKER () ^ (1ull << 63)) % (r - l + 1) + l;}
bool Memory_Begins;
const double pi = acos (-1);
//__gnu_pbds :: tree < Key, Mapped, Cmp_Fn = std :: less < Key >, Tag = rb_tree_tag, Node_Upadte = null_tree_node_update, Allocator = std :: allocator < char > > ;
//__gnu_pbds :: tree < PPS, __gnu_pbds :: null_type, less < PPS >, __gnu_pbds :: rb_tree_tag, __gnu_pbds :: tree_order_statistics_node_update > tr;
template < class Z >
inline void read (Z &tmp) {
	Z x = 0, f = 0;
	char c = getchar ();
	for ( ; c < '0' || c > '9' ; c = getchar ()) f |= (c == '-');
	for ( ; c >= '0' && c <= '9' ; c = getchar ()) x = (x << 1) + (x << 3) + (c & 15);
	tmp = !f ? x : -x;
}
const int N = 1e5 + 5;
vector < int > G[N];
int n, m, dfn[N], low[N], isin[N], time_click, scc[N], cnt, siz[N];
vector < int > sccs[N];
stack < int > stk;
inline void tarjan (int u) {
	low[u] = dfn[u] = ++ time_click;
	isin[u] = 1;
	stk.push (u);
	for (int v : G[u]) {
		if (!dfn[v]) {tarjan (v), low[u] = min (low[u], low[v]);}
		else {if (isin[v]) low[u] = min (low[u], dfn[v]);}
	}
	if (dfn[u] == low[u]) {
		cnt ++;
		while (true) {
			int v = stk.top (); scc[v] = cnt, isin[v] = 0, stk.pop ();
			if (u == v) break;
		}
	}
}
int mod;
int f[N], g[N], st[N], vis[N], deg[N];
vector < pair < int, int > > ed;
set < int > gr[N];
bool Memory_Ends;
signed main () {
	fprintf (stderr, "%.3lf MB\n", (&Memory_Begins - &Memory_Ends) / 1048576.0);
	read (n), read (m), read (mod);
	for (int i = 1;i <= m; ++ i) {
		int u, v;
		read (u), read (v);
		G[u].push_back (v);
		ed.push_back (make_pair (u, v));
	}
	for (int i = 1;i <= n; ++ i) {
		if (!scc[i]) tarjan (i);
	}
	for (int i = 1;i <= n; ++ i) {
		sccs[scc[i]].push_back (i);
		siz[scc[i]] ++;
	}
	for (int i = 1;i <= cnt; ++ i) st[i] = 1;
	for (pair < int, int > e : ed) {
		int u = e.first, v = e.second;
		u = scc[u], v = scc[v];
		if (u != v) gr[u].insert (v)/*, printf ("%d %d\n", u, v)*/;
		if (u != v) st[v] = 0;
	}
	for (int i = 1;i <= n; ++ i) {
		for (int x : gr[i]) deg[x] ++;
	}
//	printf ("ok: ");
//	for (int i = 1;i <= cnt; ++ i) {
//		if (st[i]) printf ("%d ", i);
//	}
//	printf ("\n");
	queue < int > q; 
	while (!q.empty ()) q.pop (); 
	for (int i = 1;i <= cnt; ++ i) {
		if (!deg[i]) f[i] = siz[i], g[i] = 1, q.push (i);
	}
	while (!q.empty ()) {
		int u = q.front (); q.pop ();
		if (vis[u]) continue;
		vis[u] = 1;
		for (int v : gr[u]) {
//			printf ("f[%d] -> f[%d], suc = %d\n", u, v, f[v] <= f[u] + siz[v]);
			if (f[v] > f[u] + siz[v]) ;
			else if (f[v] == f[u] + siz[v]) g[v] = (g[v] + g[u]) % mod;
			else f[v] = f[u] + siz[v], g[v] = g[u];
			deg[v] --;
			if (!deg[v]) q.push (v);
		}
	}
//	for (int i = 1;i <= n; ++ i) {
//		printf ("%d ", scc[i]);
//	}
//	printf ("\n");
//	for (int i = 1;i <= cnt; ++ i) {
//		printf ("qwq: %d %d\n", f[i], g[i]);
//	}
	int mxlen = -1, tot = 0;
	for (int i = 1;i <= cnt; ++ i) {
		if (f[i] > mxlen && g[i] > 0) mxlen = f[i], tot = g[i];
		else if (f[i] == mxlen) tot = (tot + g[i]) % mod;
	}
	printf ("%d\n%d\n", mxlen, tot);
	fprintf (stderr, "%.3lf ms\n", 1e3 * clock () / CLOCKS_PER_SEC);
	return 0;
}
/*
Things to check:
1. int ? long long ? unsigned int ? unsigned long long ?
2. array size ? is it enough ?
*/

六、洛谷 P3627 [APIO2009] 抢掠计划

https://www.luogu.com.cn/problem/P3627

首先考虑缩点，然后就可以建出一张新图，拿样例举个例子：

缩点之后 \(1, 2, 4\) 被缩成一个点，为 \(1\)：

所点之后每个点的权值（钱数）就是点所在 SCC 的钱数的总和。

然后把起点所能到的子图留下来，容易发现是一张 DAG，然后跑 dp 就行了，注意终点一定要在酒吧所在的 SCC。

注意重边 & 自环。

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/hash_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
typedef long long ll;
typedef pair < int, int > PII;
typedef int itn;
mt19937 RND_MAKER (chrono :: steady_clock :: now ().time_since_epoch ().count ());
inline ll randomly (const ll l, const ll r) {return (RND_MAKER () ^ (1ull << 63)) % (r - l + 1) + l;}
bool Memory_Begins;
const double pi = acos (-1);
//__gnu_pbds :: tree < Key, Mapped, Cmp_Fn = std :: less < Key >, Tag = rb_tree_tag, Node_Upadte = null_tree_node_update, Allocator = std :: allocator < char > > ;
//__gnu_pbds :: tree < PPS, __gnu_pbds :: null_type, less < PPS >, __gnu_pbds :: rb_tree_tag, __gnu_pbds :: tree_order_statistics_node_update > tr;
template < class Z >
inline void read (Z &tmp) {
	Z x = 0, f = 0;
	char c = getchar ();
	for ( ; c < '0' || c > '9' ; c = getchar ()) f |= (c == '-');
	for ( ; c >= '0' && c <= '9' ; c = getchar ()) x = (x << 1) + (x << 3) + (c & 15);
	tmp = !f ? x : -x;
}
const int N = 5e5 + 5;
vector < int > G[N];
int n, m, dfn[N], low[N], isin[N], time_click, scc[N], cnt;
stack < int > stk;
inline void tarjan (int u) {
	low[u] = dfn[u] = ++ time_click;
	isin[u] = 1;
	stk.push (u);
	for (int v : G[u]) {
		if (!dfn[v]) {tarjan (v), low[u] = min (low[u], low[v]);}
		else {if (isin[v]) low[u] = min (low[u], dfn[v]);}
	}
	if (dfn[u] == low[u]) {
		cnt ++;
		while (true) {
			int v = stk.top (); scc[v] = cnt, isin[v] = 0, stk.pop ();
			if (u == v) break;
		}
	}
}
ll val[N], w[N], f[N];
int deg[N];
bool used[N];
vector < int > bar;
vector < int > G2[N], F[N];
int sta;
bool Memory_Ends;
signed main () {
	fprintf (stderr, "%.3lf MB\n", (&Memory_Begins - &Memory_Ends) / 1048576.0);
	read (n), read (m);
	vector < PII > ed; ed.clear ();
	for (int i = 1;i <= m; ++ i) {
		int u, v;
		read (u), read (v);
		G[u].push_back (v);
		ed.push_back (make_pair (u, v)); 
	} 
	for (int i = 1;i <= n; ++ i) {if (!dfn[i]) tarjan (i);}
//	for (int i = 1;i <= n; ++ i) printf ("%d ", scc[i]); printf ("\n"); 
	for (int i = 1;i <= n; ++ i) read (w[i]);
	read (sta);
	int _; read (_);
	for (int i = 1;i <= _; ++ i) {
		int x; read (x);
		bar.push_back (x);
	}
	for (int i = 1;i <= n; ++ i) val[scc[i]] += 1ll * w[i];
	set < PII > st; st.clear ();
	for (PII e : ed) {
		int u = e.first, v = e.second;
		u = scc[u], v = scc[v];
		if (u ^ v) st.insert (make_pair (u, v)); 
	}
	for (PII e : st) {
//		printf ("%d %d\n", e.first, e.second);
		G2[e.first].push_back (e.second);
	}
	sta = scc[sta];
	queue < int > q;
	q.push (sta);
	for (int i = 1;i <= cnt; ++ i) deg[i] = 0;
	while (!q.empty ()) {
		int u = q.front (); q.pop ();
		if (used[u]) continue;
		used[u] = true;
		for (int v : G2[u]) {
			F[u].push_back (v);
//			printf ("%d %d\n", u, v);
			deg[v] ++;
			q.push (v);
		}
	}
	q.push (sta);
	for (int i = 1;i <= cnt; ++ i) f[i] = val[i];
	while (!q.empty ()) {
		int u = q.front (); q.pop ();
		if (deg[u]) continue;
		for (int v : F[u]) {
			deg[v] --;
			f[v] = max (f[v], f[u] + val[v]);
			if (!deg[v]) {
				q.push (v);
			}
		}
	}
	ll ans = 0;
	for (int i = 1;i <= bar.size (); ++ i) ans = max (ans, f[scc[bar[i - 1]]]);
	printf ("%lld\n", ans);
	fprintf (stderr, "%.3lf ms\n", 1e3 * clock () / CLOCKS_PER_SEC);
	return 0;
}
/*
Things to check:
1. int ? long long ? unsigned int ? unsigned long long ?
2. array size ? is it enough ?
*/

七、洛谷 P2403 [SDOI2010] 所驼门王的宝藏

https://www.luogu.com.cn/problem/P2403

样例图：

首先我们考虑建图。

将每一行有 1 类型门的节点建成一个环，然后将环上任意一点连向每一行对应剩下的节点。

每一列同理。

然后 3 号门就暴力连边即可。

建图之后跑一遍 Tarjan，然后带权求最长链即可。

我的代码不开 O2 \(80\) 分，开了 O2 才 \(100\) 分，所以仅供参考。

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/hash_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
typedef long long ll;
typedef pair < int, int > PII;
typedef int itn;
mt19937 RND_MAKER (chrono :: steady_clock :: now ().time_since_epoch ().count ());
inline ll randomly (const ll l, const ll r) {return (RND_MAKER () ^ (1ull << 63)) % (r - l + 1) + l;}
bool Memory_Begins;
const double pi = acos (-1);
//__gnu_pbds :: tree < Key, Mapped, Cmp_Fn = std :: less < Key >, Tag = rb_tree_tag, Node_Upadte = null_tree_node_update, Allocator = std :: allocator < char > > ;
//__gnu_pbds :: tree < PPS, __gnu_pbds :: null_type, less < PPS >, __gnu_pbds :: rb_tree_tag, __gnu_pbds :: tree_order_statistics_node_update > tr;
template < class Z >
inline void read (Z &tmp) {
	Z x = 0, f = 0;
	char c = getchar ();
	for ( ; c < '0' || c > '9' ; c = getchar ()) f |= (c == '-');
	for ( ; c >= '0' && c <= '9' ; c = getchar ()) x = (x << 1) + (x << 3) + (c & 15);
	tmp = !f ? x : -x;
}
const int N = 1e5 + 5;
vector < int > G[N];
int dfn[N], low[N], isin[N], time_click, scc[N], cnt;
stack < int > stk;
inline void tarjan (int u) {
	low[u] = dfn[u] = ++ time_click;
	isin[u] = 1;
	stk.push (u);
	for (int v : G[u]) {
		if (!dfn[v]) {tarjan (v), low[u] = min (low[u], low[v]);}
		else {if (isin[v]) low[u] = min (low[u], dfn[v]);}
	}
	if (dfn[u] == low[u]) {
		cnt ++;
		while (true) {
			int v = stk.top (); scc[v] = cnt, isin[v] = 0, stk.pop ();
			if (u == v) break;
		}
	}
}
int n, r, c;
int room[N][3], val[N], dp[N], deg[N];
vector < int > type[1000005], o[1000005];
vector < int > gr[N];
map < pair < int, int >, int > mp;
vector < pair < int, int > > ed;
bool Memory_Ends;
signed main () {
	fprintf (stderr, "%.3lf MB\n", (&Memory_Begins - &Memory_Ends) / 1048576.0);
	read (n), read (r), read (c);
	mp.clear ();
	for (int i = 1;i <= n; ++ i) {
		read (room[i][0]), read (room[i][1]), read (room[i][2]);
		mp[make_pair (room[i][0], room[i][1])] = i;
	}
	for (int i = 1;i <= n; ++ i) {
		if (room[i][2] == 1) type[room[i][0]].push_back (i);
		if (room[i][2] != 1) o[room[i][0]].push_back (i);
	}
	for (int i = 1;i <= r; ++ i) {
		for (int j = 0;j < type[i].size (); ++ j) {
			if (j + 1 == type[i].size ()) G[type[i][j]].push_back (type[i][0]);
			else G[type[i][j]].push_back (type[i][j + 1]);
		}
		if (type[i].empty ()) continue;
		int st = type[i][0];
		for (int u : o[i]) {
			G[st].push_back (u);
		}
	}
	for (int i = 1;i <= n; ++ i) {
		if (room[i][2] == 1) type[room[i][0]].pop_back ();
		if (room[i][2] != 1) o[room[i][0]].pop_back ();
	}	
	for (int i = 1;i <= n; ++ i) {
		if (room[i][2] == 2) type[room[i][1]].push_back (i);
		if (room[i][2] != 2) o[room[i][1]].push_back (i);
	}
	for (int i = 1;i <= c; ++ i) {
		for (int j = 0;j < type[i].size (); ++ j) {
			if (j + 1 == type[i].size ()) G[type[i][j]].push_back (type[i][0]);
			else G[type[i][j]].push_back (type[i][j + 1]); 
		}
		if (type[i].empty ()) continue;
		int st = type[i][0];
		for (int u : o[i]) {
			G[st].push_back (u);
		}
	}
	for (int i = 1;i <= n; ++ i) {
		if (room[i][2] == 3) {
			for (int j = -1;j <= 1; ++ j) {
				for (int k = -1;k <= 1; ++ k) {
					if (!j && !k) continue;
					int _ = mp[make_pair (room[i][0] + j, room[i][1] + k)];
					G[i].push_back (_);
				}
			}
		}
	}
	for (int i = 1;i <= n; ++ i) {
		if (!scc[i]) tarjan (i);
	}
	for (int i = 1;i <= n; ++ i) val[scc[i]] ++;
//	set < pair < int, int > > ed;
	ed.clear ();
	for (int i = 1;i <= n; ++ i) {
		for (int v : G[i]) {
			int u = i;
			u = scc[u];
			int V = scc[v];
			if (u != V) ed.push_back (make_pair (u, V)); 
		}
	}
	for (pair < int, int > e : ed) {
		gr[e.first].push_back (e.second);
		deg[e.second] ++; 
	}
	queue < int > q;
	while (!q.empty ()) q.pop ();
	for (int i = 1;i <= cnt; ++ i) {
		if (!deg[i]) q.push (i);
		dp[i] = val[i]; 
	}
	while (!q.empty ()) {
		int u = q.front (); q.pop ();
		for (int v : gr[u]) {
			dp[v] = max (dp[v], dp[u] + val[v]);
			deg[v] --;
			if (!deg[v]) q.push (v);
		}
	}
	int ans = 0;
	for (int i = 1;i <= cnt; ++ i) ans = max (ans, dp[i]);
	printf ("%d\n", ans);
	fprintf (stderr, "%.3lf ms\n", 1e3 * clock () / CLOCKS_PER_SEC);
	return 0;
}
/*
Things to check:
1. int ? long long ? unsigned int ? unsigned long long ?
2. array size ? is it enough ?
*/

写在最后：做题时遇到的坑点

缩点后一定要判重边 & 自环！！1

posted @ 2023-06-23 22:00 CountingGroup 阅读(17) 评论(0) 编辑收藏举报

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