LCA倍增 模板
2018-08-08 本文已影响21人
JesHrz
LCA倍增 最近公共祖先
构造 NlogN 查询 ogN
先调用pre()
构造对数数组 再调用dfs(root, 0, 0)
查询深度 再调用work()
构造跳表
最后调用lca(u, v)
查询在有根树中节点u和v的最近公共祖先
const int N = 10005;
struct edge { int to, next; } e[N << 1];
int n, head[N];
namespace LCA
{
int log[N], depth[N], deg[N], par[N][30];
inline void pre()
{
log[1] = 0;
for (int i = 2; i < N; ++i)
{
log[i] = log[i - 1];
if ((1 << (log[i] + 1)) == i) log[i]++;
}
}
void dfs(int u, int fa, int dep)
{
depth[u] = dep;
par[u][0] = fa;
for (int i = head[u]; i; i = e[i].next)
{
int v = e[i].to;
if (v == fa) continue;
dfs(v, u, dep + 1);
}
}
inline void work()
{
for (int j = 1; j <= log[n]; ++j)
for (int i = 1; i <= n; ++i)
par[i][j] = par[par[i][j - 1]][j - 1];
}
inline int lca(int u, int v)
{
if (depth[u] < depth[v]) std::swap(u, v);
int t = depth[u] - depth[v];
for (int j = 0; j <= log[n]; ++j)
if (t >> j & 1) u = par[u][j];
if (u == v) return u;
for (int i = log[n]; ~i; --i)
if (par[u][i] != par[v][i])
{
u = par[u][i];
v = par[v][i];
}
return par[u][0];
}
}