Solution -「洛谷 P4320」道路相遇

\(\mathcal{Description}\)

  Link.

  给定一个 \(n\) 个点 \(m\) 条边的连通无向图，并给出 \(q\) 个点对 \((u,v)\)，询问 \(u\) 到 \(v\) 的路径所必经的结点个数。

  \(n,q\le5\times10^5\)，\(q\le\min\{\frac{n(n-1)}2,10^6\}\)。

\(\mathcal{Solution}\)

  大概是双倍经验吧。

  建出圆方树，预处理圆方树上每个点到根经过的圆点个数，然后求 LCA 计算答案即可。

  复杂度 \(\mathcal O(n\log n)-\mathcal O(\log n)\)。

\(\mathcal{Code}\)

#include <cstdio>

const int MAXN = 5e5, MAXM = 1e6;
int n, m, q, snode;
int dfc, top, dfn[MAXN + 5], low[MAXN + 5], stk[MAXN + 5];
int dep[MAXN * 2 + 5], cnt[MAXN * 2], fa[MAXN * 2 + 5][20];

struct Graph {
	int ecnt, head[MAXN * 2 + 5], to[MAXM * 2 + 5], nxt[MAXM * 2 + 5];
	inline void link ( const int s, const int t ) {
		to[++ ecnt] = t, nxt[ecnt] = head[s];
		head[s] = ecnt;
	}
	inline void add ( const int u, const int v ) {
		link ( u, v ), link ( v, u );
	}
} src, tre;

inline bool chkmin ( int& a, const int b ) { return b < a ? a = b, true : false; }

inline void Tarjan ( const int u, const int f ) {
	dfn[u] = low[u] = ++ dfc, stk[++ top] = u;
	for ( int i = src.head[u], v; i; i = src.nxt[i] ) {
		if ( ( v = src.to[i] ) == f ) continue;
		if ( ! dfn[v] ) {
			Tarjan ( v, u ), chkmin ( low[u], low[v] );
			if ( low[v] >= dfn[u] ) {
				tre.add ( u, ++ snode );
				do tre.add ( snode, stk[top] ); while ( stk[top --] ^ v );
			}
		} else chkmin ( low[u], dfn[v] );
	}
}

inline void init ( const int u, const int f ) {
	dep[u] = dep[fa[u][0] = f] + 1, cnt[u] = cnt[f] + ( u <= n );
	for ( int i = 1; i <= 19; ++ i ) fa[u][i] = fa[fa[u][i - 1]][i - 1];
	for ( int i = tre.head[u], v; i; i = tre.nxt[i] ) {
		if ( ( v = tre.to[i] ) ^ f ) {
			init ( v, u );
		}
	}
}

inline int calcLCA ( int u, int v ) {
	if ( dep[u] < dep[v] ) u ^= v ^= u ^= v;
	for ( int i = 19; ~ i; -- i ) if ( dep[fa[u][i]] >= dep[v] ) u = fa[u][i];
	if ( u == v ) return u;
	for ( int i = 19; ~ i; -- i ) if ( fa[u][i] ^ fa[v][i] ) u = fa[u][i], v = fa[v][i];
	return fa[u][0];
}

int main () {
	scanf ( "%d %d", &n, &m ), snode = n;
	for ( int i = 1, u, v; i <= m; ++ i ) {
		scanf ( "%d %d", &u, &v );
		src.add ( u, v );
	}
	Tarjan ( 1, 0 ), init ( 1, 0 );
	scanf ( "%d", &q );
	for ( int i = 1, u, v; i <= q; ++ i ) {
		scanf ( "%d %d", &u, &v );
		int w = calcLCA ( u, v );
		printf ( "%d\n", cnt[u] + cnt[v] - cnt[w] - cnt[fa[w][0]] );
	}
	return 0;
}