\(\mathcal{Description}\)
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;
}