求最近公共祖先LCA兩種方法
阿新 • • 發佈:2020-07-10
tarjan求lca
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這種演算法本質上是用並查集對向上標記法的優化,是離線演算法,即一次性讀入所有詢問,統一計算,統一輸出。
時間複雜度\(O(n+m)\)v[]
進行標記
\(v[x]\doteq 0\) --> x節點未訪問過
\(v[x]\doteq 1\) --> x節點已經訪問,但未回溯
\(v[x]\doteq 2\) --> x節點已經訪問並回溯
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code
#include <cstdio> #include <vector> #include <algorithm> using namespace std; const int N = 1e5+5; struct side { int t, d, next; }e[N<<1]; int head[N], tot; void add(int x, int y, int z) { e[++tot].next = head[x]; head[x] = tot; e[tot].t = y, e[tot].d = z; } int n, m, Q, d[N], f[N], v[N], ans[N]; vector<int> q[N], h[N]; int found(int x) { return x == f[x] ? x : (f[x] = found(f[x])); } void tarjan(int x) { v[x] = 1; for (int i = head[x]; i; i = e[i].next) { int y = e[i].t; if (v[y]) continue; d[y] = d[x] + e[i].d; tarjan(y); f[y] = x; } for (int i = 0; i < q[x].size(); i++) { int y = q[x][i], id = h[x][i]; if (v[y] != 2) continue; ans[id] = min(ans[id], d[x] + d[y] - 2*d[found(y)]); } v[x] = 2; } int main() { scanf("%d%d", &n, &m); for (int i = 1; i <= n; i++) f[i] = i; while (m--) { int x, y, z; scanf("%d%d%d", &x, &y, &z); add(x, y, z); add(y, x, z); } scanf("%d", &Q); for (int i = 1; i <= Q; i++) { int x, y; scanf("%d%d", &x, &y); if (x == y) continue; ans[i] = 1 << 30; q[x].push_back(y), h[x].push_back(i); q[y].push_back(x), h[y].push_back(i); } tarjan(1); for (int i = 1; i <= Q; i++) printf("%d\n", ans[i]); return 0; }
倍增求Lca
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f[x][k]表示從x向根節點走\(2^k\)步到達的節點
d[x]表示樹的深度
預處理時間複雜度為\(O(n\log n)\),每次詢問時間複雜度為\(O(\log n)\)code
int d[N], f[N][21]; void dfs(int x, int fa) { f[x][0] = fa; d[x] = d[fa] + 1; for (int i = 1; i <= 20; i++) f[x][i] = f[f[x][i-1]][i-1]; for (int i = head[x]; i; i = e[i].next) if (e[i].t != fa) dfs(e[i].t, x); } int lca(int x, int y) { if (d[x] > d[y]) swap(x, y); y = found(y, d[y] - d[x]); if (x == y) return x; for (int i = 20; i >= 0; i--) if (f[x][i] != f[y][i]) x = f[x][i], y = f[y][i]; return f[x][0]; }