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Step1 Problem:
[原题] 给定一棵有根多叉树，请求出指定两个点直接最近的公共祖先。
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Step2 Ideas:
lca模板题，主要为了存模板。LCA（Least Common Ancestors），即最近公共祖先，是指在有根树中，找出某两个结点u和v最近的公共祖先
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Step3 Code: Tarjan，离线算法，时间复杂度O(n + q)
1 // luogu-judger-enable-o2
2 #define _CRT_SECURE_NO_WARNINGS
3 #include<cstdio>
4 #include<iostream>
5 #include<cstring>
6 #include<algorithm>
7 #include<bitset>
8 #include<cassert>
9 #include<cctype>
10 #include<math.h>
11 #include<cstdlib>
12 #include<ctime>
13 #include<deque>
14 #include<iomanip>
15 #include<list>
16 #include<map>
17 #include<queue>
18 #include<set>
19 #include<stack>
20 #include<vector>
21 #define lt k<<1
22 #define rt k<<1|1
23 #define lowbit(x) x&(-x)
24 #define lson l,mid,lt
25 #define rson mid+1,r,rt
26 using namespace std;
27 typedef long long ll;
28 typedef long double ld;
29 #define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
30 #define mem(a, b) memset(a, b, sizeof(a))
31 //#define int ll
32 const double pi = acos(-1.0);
33 const double eps = 1e-6;
34 const double C = 0.57721566490153286060651209;
35 const ll mod = 3e7;
36 const int inf = 0x3f3f3f3f;
37 const ll INF = 0x3f3f3f3f3f3f3f3f;
38 const int maxn = 5e5+ 4000;
39 struct node {
40 int u, v, w, next;
41 node () {}
42 node(int u_, int v_, int w_, int next_) {
43 u = u_;
44 v = v_;
45 w = w_;
46 next = next_;
47 }
48 } edge[maxn << 1];
49
50 struct nod {
51 int u, v, id, next;
52 nod() {}
53 nod(int u_, int v_, int id_, int next_) {
54 u = u_;
55 v = v_;
56 id = id_;
57 next = next_;
58 }
59 }qu[maxn << 1];
60 int n, m, ans;
61 int head[maxn], head1[maxn];
62 int cnt, cnt1;
63 int res[maxn];
64 int de[maxn], pre[maxn], fa[maxn];
65 bool vis[maxn];
66
67 void init() {
68 cnt = cnt1 = 0;
69 mem(vis, false);
70 mem(de, 0);
71 mem(head, -1);
72 mem(head1, -1);
73 mem(qu, 0);
74 mem(edge, 0);
75 for (int i = 1; i <= n; i++) fa[i] = i;
76 }
77
78 void add_edge(int u, int v, int w) {
79 edge[cnt] = (node){ u, v, w, head[u] };
80 head[u] = cnt++;
81 }
82
83 void add_qu(int u, int v, int id) {
84 qu[cnt1] = (nod){ u, v, id, head1[u] };
85 head1[u] = cnt1++;
86 }
87
88 int find_fa(int x) {
89 if (x == fa[x]) return x;
90 else return fa[x] = find_fa(fa[x]);
91 }
92
93 void Tarjan(int u) {
94 vis[u] = true;
95 for (int i = head[u]; ~i; i = edge[i].next) {
96 int v = edge[i].v;
97 if (!vis[v]) {
98 Tarjan(v);
99 int r1 = find_fa(u);
100 int r2 = find_fa(v);
101 if (r1 != r2) fa[r2] = r1;
102 }
103 }
104 for (int i = head1[u]; ~i; i = qu[i].next) {
105 int v = qu[i].v;
106 if (vis[v]) res[qu[i].id] = find_fa(v);
107 }
108 }
109
110 int main() {
111 // freopen("testdata.in", "r", stdin);
112 int s;
113 cin >> n >> m >> s;
114 init();
115 for (int i = 1; i < n; i++) {
116 int u, v;
117 cin >> u >> v;
118 add_edge(u, v, 1);
119 add_edge(v, u, 1);
120 }
121 for (int i = 1; i <= m; i++) {
122 int a, b;
123 cin >> a >> b;
124 add_qu(a, b, i);
125 add_qu(b, a, i);
126 }
127 Tarjan(s);
128 for (int i = 1; i <= m; i++) cout << res[i] << endl;
129 return 0;
130 }
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Step4 Code2: 倍增法求lca，在线算法，时间复杂度O(n logn)
1 #define _CRT_SECURE_NO_WARNINGS
2 #include<cstdio>
3 #include<iostream>
4 #include<cstring>
5 #include<algorithm>
6 #include<bitset>
7 #include<cassert>
8 #include<cctype>
9 #include<math.h>
10 #include<cstdlib>
11 #include<ctime>
12 #include<deque>
13 #include<iomanip>
14 #include<list>
15 #include<map>
16 #include<queue>
17 #include<set>
18 #include<stack>
19 #include<vector>
20 #define lt k<<1
21 #define rt k<<1|1
22 #define lowbit(x) x&(-x)
23 #define lson l,mid,lt
24 #define rson mid+1,r,rt
25 using namespace std;
26 typedef long long ll;
27 typedef long double ld;
28 #define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
29 #define mem(a, b) memset(a, b, sizeof(a))
30 //#define int ll
31 const double pi = acos(-1.0);
32 const double eps = 1e-6;
33 const double C = 0.57721566490153286060651209;
34 const ll mod = 3e7;
35 const int inf = 0x3f3f3f3f;
36 const ll INF = 0x3f3f3f3f3f3f3f3f;
37 const int maxn = 5e5+ 4000;
38 int lg[maxn], dep[maxn], fa[maxn][20], head[maxn];
39 int cnt;
40 int n, m, s;
41
42 struct node {
43 int v, next;
44 } edge[maxn << 1];
45
46 void add_edge(int u, int v) {
47 edge[cnt].v = v;
48 edge[cnt].next = head[u];
49 head[u] = cnt++;
50 }
51
52 void init() {
53 cnt = 0;
54 for (int i = 1; i <= n; i++) {
55 head[i] = -1;
56 dep[i] = -1;
57 lg[i] = lg[i - 1] + (1 << (lg[i - 1]) == i);
58 }
59 mem(fa, 0);
60 }
61
62 void dfs(int u, int fath) {
63 dep[u] = dep[fath] + 1;
64 fa[u][0] = fath;
65 for (int i = 1; (1 << i) <= dep[u]; i++) {
66 fa[u][i] = fa[fa[u][i - 1]][i - 1];
67 }
68 for (int i = head[u]; ~i; i = edge[i].next) {
69 if (edge[i].v != fath) dfs(edge[i].v, u);
70 }
71 }
72
73 int lca(int u, int v) {
74 if (dep[u] < dep[v]) swap(u, v);
75 while (dep[u] > dep[v]) u = fa[u][lg[dep[u] - dep[v]] - 1];
76 if (u == v) return u;
77 for (int i = lg[dep[u]] - 1; i >= 0; i--) {
78 if (fa[u][i] != fa[v][i]) {
79 u = fa[u][i];
80 v = fa[v][i];
81 }
82 }
83 return fa[u][0];
84 }
85
86 int main() {
87 scanf("%d%d%d", &n, &m, &s);
88 init();
89 for (int i = 1; i < n; i++) {
90 int u, v;
91 cin >> u >> v;
92 add_edge(u, v);
93 add_edge(v, u);
94 }
95 dfs(s, 0);
96 for (int i = 1; i <= m; i++) {
97 int u, v;
98 scanf("%d%d", &u, &v);
99 printf("%d\n", lca(u, v));
100 }
101 return 0;
102 }