洛谷上的lca模板题——传送门
1.tarjan求lca
学了求lca的tarjan算法（离线），在洛谷上做模板题，结果后三个点超时。
又把询问改成链式前向星，才ok。
这个博客，tarjan分析的很详细。
附代码——
#include <cstdio>
#include <cstring>
const int maxn = 500001;
int n, m, cnt, s, cns;
int x, y, z[maxn];//z是x和y的lca
int f[maxn], head[maxn], from[maxn];
bool vis[maxn];
struct node
{
int to, next;
}e[2 * maxn];
struct Node
{
int to, next, num;
}q[2 * maxn];
inline int read()//读入优化
{
int x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-') f = -1;
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
inline void ask(int u, int v, int i)//储存待询问的结构体，也是链式前向星优化
{
q[cns].num = i;//num表示第几次询问
q[cns].to = v;
q[cns].next = from[u];
from[u] = cns++;
}
inline void add(int u, int v)//
{
e[cnt].to = v;
e[cnt].next = head[u];
head[u] = cnt++;
}
inline int find(int a)
{
return a == f[a] ? a : f[a] = find(f[a]);//路径压缩优化
}
/*inline void Union(int a, int b)
{
int fx = find(a), fy = find(b);
if(fx == fy) return;
f[fy] = fx;
}*/
inline void tarjan(int k)
{
int i, j;
vis[k] = 1;
f[k] = k;
for(i = head[k]; i != -1; i = e[i].next)
if(!vis[e[i].to])
{
tarjan(e[i].to);
//Union(k, e[i].to);
f[e[i].to] = k;
}
for(i = from[k]; i != -1; i = q[i].next)
if(vis[q[i].to] == 1)
z[q[i].num] = find(q[i].to);
}
int main()
{
int i, j, u, v;
n = read();
m = read();
s = read();
memset(head, -1, sizeof(head));
memset(from, -1, sizeof(from));
for(i = 1; i <= n - 1; i++)
{
u = read();
v = read();
add(u, v);//注意添加两遍
add(v, u);
}
for(i = 1; i <= m; i++)
{
x = read();
y = read();
ask(x, y, i);//两遍
ask(y, x, i);
}
tarjan(s);
for(i = 1; i <= m; i++) printf("%d\n", z[i]);
return 0;
}
进过培训，修改了代码
1 # include <iostream>
2 # include <cstdio>
3 # include <cstring>
4 # include <string>
5 # include <cmath>
6 # include <vector>
7 # include <map>
8 # include <queue>
9 # include <cstdlib>
10 # define MAXN 500001
11 using namespace std;
12
13 inline int get_num() {
14 int k = 0, f = 1;
15 char c = getchar();
16 for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
17 for(; isdigit(c); c = getchar()) k = k * 10 + c - '0';
18 return k * f;
19 }
20
21 int n, m, s;
22 int fa[MAXN], qx[MAXN], qy[MAXN], ans[MAXN], f[MAXN];
23 vector <int> vec[MAXN], q[MAXN];
24
25 inline int find(int x)
26 {
27 return x == fa[x] ? x : fa[x] = find(fa[x]);
28 }
29
30 inline void dfs(int u)
31 {
32 int i, v;
33 fa[u] = u;
34 for(i = 0; i < vec[u].size(); i++)
35 {
36 v = vec[u][i];
37 if(f[u] != v) f[v] = u, dfs(v);
38 }
39 for(i = 0; i < q[u].size(); i++)
40 if(f[v = u ^ qx[q[u][i]] ^ qy[q[u][i]]])
41 ans[q[u][i]] = find(v);
42 fa[u] = f[u];
43 }
44
45 int main()
46 {
47 int i, x, y;
48 n = get_num();
49 m = get_num();
50 s = get_num();
51 for(i = 1; i < n; i++)
52 {
53 x = get_num();
54 y = get_num();
55 vec[x].push_back(y);
56 vec[y].push_back(x);
57 }
58 for(i = 1; i <= m; i++)
59 {
60 qx[i] = get_num();
61 qy[i] = get_num();
62 q[qx[i]].push_back(i);
63 q[qy[i]].push_back(i);
64 }
65 dfs(s);
66 for(i = 1; i <= m; i++) printf("%d\n", ans[i]);
67 return 0;
68 }
其实上面两个代码有些重复运算，请手动把求lca的过程放到dfs上面（也就是遍历到这个节点就求lca，而不是遍历完再求）
2.倍增求lca
下面是求lca的倍增算法（在线）。
1. DFS预处理出所有节点的深度和父节点
inline void dfs(int u)
{
int i;
for(i=head[u];i!=-1;i=next[i])
{
if (!deep[to[i]])
{
deep[to[i]] = deep[u]+1;
p[to[i]][0] = u; //p[x][0]保存x的父节点为u;
dfs(to[i]);
}
}
}
2. 初始各个点的2^j祖先是谁 ,其中 2^j (j =0...log(该点深度))倍祖先，1倍祖先就是父亲，2倍祖先是父亲的父亲......。
void init()
{
int i,j;
//p[i][j]表示i结点的第2^j祖先
for(j=1;(1<<j)<=n;j++)
for(i=1;i<=n;i++)
if(p[i][j-1]!=-1)
p[i][j]=p[p[i][j-1]][j-1];//i的第2^j祖先就是i的第2^(j-1)祖先的第2^(j-1)祖先
}
3.从深度大的节点上升至深度小的节点同层，如果此时两节点相同直接返回此节点，即lca。
否则，利用倍增法找到最小深度的 p[a][j]!=p[b][j]，此时他们的父亲p[a][0]即lca。
int lca(int a,int b)//最近公共祖先
{
int i,j;
if(deep[a]<deep[b])swap(a,b);
for(i=0;(1<<i)<=deep[a];i++);
i--;
//使a，b两点的深度相同
for(j=i;j>=0;j--)
if(deep[a]-(1<<j)>=deep[b])
a=p[a][j];
if(a==b)return a;
//倍增法，每次向上进深度2^j，找到最近公共祖先的子结点
for(j=i;j>=0;j--)
{
if(p[a][j]!=-1&&p[a][j]!=p[b][j])
{
a=p[a][j];
b=p[b][j];
}
}
return p[a][0];
}
最后是完整代码，为了节约时间，就没有把p数组初始化为-1.
#include <cstdio>
#include <cstring>
#include <iostream>
const int maxn = 500001;
int n, m, cnt, s;
int next[2 * maxn], to[2 * maxn], head[2 * maxn], deep[maxn], p[maxn][21];
inline void add(int x, int y)
{
to[cnt] = y;
next[cnt] = head[x];
head[x] = cnt++;
}
inline void dfs(int i)
{
int j;
for(j = head[i]; j != -1; j = next[j])
if(!deep[to[j]])
{
deep[to[j]] = deep[i] + 1;
p[to[j]][0] = i;
dfs(to[j]);
}
}
inline void init()
{
int i, j;
for(j = 1; (1 << j) <= n; j++)
for(i = 1; i <= n; i++)
p[i][j] = p[p[i][j - 1]][j - 1];
}
inline int lca(int a, int b)
{
int i, j;
if(deep[a] < deep[b]) std::swap(a, b);
for(i = 0; (1 << i) <= deep[a]; i++);
i--;
for(j = i; j >= 0; j--)
if(deep[a] - (1 << j) >= deep[b])
a = p[a][j];
if(a == b) return a;
for(j = i; j >= 0; j--)
if(p[a][j] != p[b][j])
{
a = p[a][j];
b = p[b][j];
}
return p[a][0];
}
int main()
{
int i, j, x, y;
memset(head, -1, sizeof(head));
scanf("%d %d %d", &n, &m, &s);
for(i = 1; i <= n - 1; i++)
{
scanf("%d %d", &x, &y);
add(x, y);
add(y, x);
}
deep[s] = 1;
dfs(s);
init();
for(i = 1; i <= m; i++)
{
scanf("%d %d", &x, &y);
printf("%d\n", lca(x, y));
}
return 0;
}
经过培训，又改了改代码。
1 # include <iostream>
2 # include <cstdio>
3 # include <cstring>
4 # include <string>
5 # include <cmath>
6 # include <vector>
7 # include <map>
8 # include <queue>
9 # include <cstdlib>
10 # define MAXN 500001
11 using namespace std;
12
13 inline int get_num() {
14 int k = 0, f = 1;
15 char c = getchar();
16 for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
17 for(; isdigit(c); c = getchar()) k = k * 10 + c - '0';
18 return k * f;
19 }
20
21 int n, m, s;
22 int f[MAXN][25], deep[MAXN];
23 vector <int> vec[MAXN];
24
25 inline void dfs(int u)
26 {
27 int i, v;
28 deep[u] = deep[f[u][0]] + 1;
29 for(i = 0; f[u][i]; i++) f[u][i + 1] = f[f[u][i]][i];
30 for(i = 0; i < vec[u].size(); i++)
31 {
32 v = vec[u][i];
33 if(!deep[v]) f[v][0] = u, dfs(v);
34 }
35 }
36
37 inline int lca(int x, int y)
38 {
39 int i;
40 if(deep[x] < deep[y]) swap(x, y);
41 for(i = 20; i >= 0; i--)
42 if(deep[f[x][i]] >= deep[y])
43 x = f[x][i];
44 if(x == y) return x;
45 for(i = 20; i >= 0; i--)
46 if(f[x][i] != f[y][i])
47 x = f[x][i], y = f[y][i];
48 return f[x][0];
49 }
50
51 int main()
52 {
53 int i, x, y;
54 n = get_num();
55 m = get_num();
56 s = get_num();
57 for(i = 1; i < n; i++)
58 {
59 x = get_num();
60 y = get_num();
61 vec[x].push_back(y);
62 vec[y].push_back(x);
63 }
64 dfs(s);
65 for(i = 1; i <= m; i++)
66 {
67 scanf("%d %d", &x, &y);
68 printf("%d\n", lca(x, y));
69 }
70 return 0;
71 }
3.树剖法求lca
1 # include <iostream>
2 # include <cstdio>
3 # include <cstring>
4 # include <string>
5 # include <cmath>
6 # include <vector>
7 # include <map>
8 # include <queue>
9 # include <cstdlib>
10 # define MAXN 500001
11 using namespace std;
12
13 inline int get_num() {
14 int k = 0, f = 1;
15 char c = getchar();
16 for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
17 for(; isdigit(c); c = getchar()) k = k * 10 + c - '0';
18 return k * f;
19 }
20
21 int n, m, s;
22 int f[MAXN], size[MAXN], top[MAXN], son[MAXN], deep[MAXN];
23 vector <int> vec[MAXN];
24
25 inline void dfs1(int u)
26 {
27 int i, v;
28 size[u] = 1;
29 deep[u] = deep[f[u]] + 1;
30 for(i = 0; i < vec[u].size(); i++)
31 {
32 v = vec[u][i];
33 if(!deep[v])
34 {
35 f[v] = u;
36 dfs1(v);
37 size[u] += size[v];
38 if(size[son[u]] < size[v]) son[u] = v;
39 }
40 }
41 }
42
43 inline void dfs2(int u, int tp)
44 {
45 int i, v;
46 top[u] = tp;
47 if(!son[u]) return;
48 dfs2(son[u], tp);
49 for(i = 0; i < vec[u].size(); i++)
50 {
51 v = vec[u][i];
52 if(v != son[u] && v != f[u]) dfs2(v, v);
53 }
54 }
55
56 inline int lca(int x, int y)
57 {
58 while(top[x] != top[y])
59 {
60 if(deep[top[x]] < deep[top[y]]) swap(x, y);
61 x = f[top[x]];
62 }
63 if(deep[x] > deep[y]) swap(x, y);
64 return x;
65 }
66
67 int main()
68 {
69 int i, x, y;
70 n = get_num();
71 m = get_num();
72 s = get_num();
73 for(i = 1; i < n; i++)
74 {
75 x = get_num();
76 y = get_num();
77 vec[x].push_back(y);
78 vec[y].push_back(x);
79 }
80 dfs1(s);
81 dfs2(s, s);
82 for(i = 1; i <= m; i++)
83 {
84 x = get_num();
85 y = get_num();
86 printf("%d\n", lca(x, y));
87 }
88 return 0;
89 }
来源：https://www.cnblogs.com/zhenghaotian/p/6658105.html