LCA（倍增，RMQ，Tarjan）
LCA（Least Common Ancestors）：最近公共祖先
题目：https://www.luogu.org/problemnew/show/P3379
倍增
先预处理出每个结点向上跳2^x层的祖先和每个结点的深度
类似快速幂，拆分deep[u] - deep[v](假设deep[u] > deep[v])，每次使u向上跳2^x步，使deep[u] = deep[v]
然后再一起往上跳，dis = deep[lca(u,v)] - deep[u]，由于dis我们不知道具体多少并且lca后面的结点都是u，v的公共祖先，所以不能像上面的一样直接拆分，于是换个思路，我们去找的最远的非公共祖先，最后肯定是到lca的子节点
比如dis = 10100（2）
从大到小枚举2^n次方，2^4前面的明显就是公共祖先
2^4时发现这时候不是公共祖先，于是先跳到这里，dis = 100（2）
同理下一步之后dis = 11（2），最后dis = 1（2）循环结束
然后lca就等于这时候的v（u = v）的直接祖先
#include
using namespace std;
const int maxn = 5e5 + 50;
const int maxlog = 20;
int n, q, root;
int par[maxlog + 5][maxn];
int dep[maxn];
int head[maxn], edgecnt;
struct P
{
int to, next;
}edge[2 * maxn];
void dfs(int u, int fa)
{
par[0][u] = fa;
dep[u] = dep[fa] + 1;
for(int i = head[u]; ~i; i = edge[i].next)
{
if(edge[i].to == fa) continue;
dfs(edge[i].to, u);
}
}
void init()
{
dfs(root, -1);
for(int k = 0; k + 1 < maxlog; k++)
{
for(int u = 1; u <= n; u++)
{
if(par[k][u] == -1)
par[k + 1][u] = -1;
else
par[k + 1][u] = par[k][par[k][u]];
}
}
}
int lca(int u, int v)
{
if(dep[u] > dep[v])
{
swap(u, v);
}
for(int i = 20; ~i; i--)
{
if(dep[par[i][v]] >= dep[u]) v = par[i][v];
}
if(v == u) return v;
for(int i = maxlog; ~i; i--)
{
if(par[i][u] != par[i][v])
{
u = par[i][u];
v = par[i][v];
}
}
return par[0][u];
}
void add(int u, int v)
{
edge[edgecnt].to = v;
edge[edgecnt].next = head[u];
head[u] = edgecnt++;
}
int main()
{
ios_base::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
cin >> n >> q >> root;
memset(head, -1, sizeof(head));
int u, v;
for(int i = 0; i < n - 1; i++)
{
cin >> u >> v;
add(u, v);
add(v, u);
}
init();
while(q--)
{
cin >> u >> v;
printf("%d\n", lca(u, v));
}
return 0;
}
基于RMQ（ST表）的算法（对于LCA ±1RMQ也可以做）
vis[i]为第i次dfs所到达的结点
dep[i]为vis[i]的深度
id[u]为u第一次被dfs时的vis编号
在dfs序中易知lca(u, v) = vis[{i, id[u] <= i <= id[v]}中dep最小的i]
上面放的模板题有点卡常数，所以放了快速read
#include
using namespace std;
const int maxn = 10e5 + 50;
const int maxlog = 20;
int n, q, root, cnt;
int head[maxn], edgecnt;
int dp[maxn][maxlog];
int rec[maxn][maxlog];
int dep[maxn];
int vis[maxn];
int id[maxn];
struct P
{
int to, next;
}edge[maxn];
void add(int u, int v)
{
edge[edgecnt].to = v;
edge[edgecnt].next = head[u];
head[u] = edgecnt++;
}
void DFS(int u, int fa, int d, int &cnt)
{
dep[cnt] = d;
vis[cnt] = u;
id[u] = cnt++;
for(int i = head[u]; ~i; i = edge[i].next)
{
int v = edge[i].to;
if(v != fa)
{
DFS(v, u, d + 1, cnt);
dep[cnt] = d;
vis[cnt++] = u;
}
}
}
void ST()
{
for(int i = 0; i < cnt; i++)
dp[i][0] = dep[i], rec[i][0] = vis[i];
for(int j = 1; (1 << j) <= cnt; j++)
{
for(int i = 0; i + (1 << j) - 1 < cnt; i++)
{
if(dp[i][j - 1] > dp[i + (1 << (j - 1))][j - 1])
dp[i][j] = dp[i + (1 << (j - 1))][j - 1], rec[i][j] = rec[i + (1 << (j - 1))][j - 1];
else
dp[i][j] = dp[i][j - 1], rec[i][j] = rec[i][j - 1];
}
}
}
void init()
{
cnt = 0;
DFS(root, -1, 0, cnt);
ST();
}
inline int RMQ(int l, int r)
{
int k = 0;
while(1 << (k + 1) <= r - l + 1) k++;
if(dp[l][k] > dp[r - (1 << k) + 1][k])
return rec[r - (1 << k) + 1][k];
else
return rec[l][k];
}
inline int read()
{
int x=0,flag=0;
char ch=getchar();
if(ch=='-') flag=1;
while(ch<'0'||ch>'9')ch=getchar();
while(ch>='0'&&ch<='9')x*=10,x+=ch-'0',ch=getchar();
if(flag) return -x;
return x;
}
int main()
{
//ios_base::sync_with_stdio(0);
//cin.tie(0); cout.tie(0);
n=read(),q=read(),root=read();
//cin >> n >> q >> root;
memset(head, -1, sizeof(head));
int u, v;
for(int i = 0; i < n - 1; i++)
{
u = read(), v = read();
// cin >> u >> v;
add(u, v);
add(v, u);
}
init();
int l, r;
while(q--)
{
l = read(), r = read();
// cin >> l >> r;
l = id[l], r = id[r];
if(l > r) swap(l, r);
printf("%d\n", RMQ(l, r));
//cout << RMQ(l, r) << endl;
}
return 0;
}
Tarjan
离线算法
在dfs时， 对于查询u，v来说，若遍历到u时发现v已经遍历过了，这时lca（u，v）就是用并查集维护的集合的root
注意合并的时候将父节点作为并查集的上级即u->v合并时par[v] = u
#include
using namespace std;
const int maxn = 1e6 + 50;
int n, q, root;
int head[maxn], edgecnt;
int qhead[maxn], quecnt;
int par[maxn];
int ans[maxn];
int vis[maxn];
struct P
{
int to, next;
}edge[maxn];
struct P1
{
int to, num, lca, next;
}que[maxn];
int Find(int u)
{
if(u == par[u])
return u;
return par[u] = Find(par[u]);
}
void Merge(int x, int y)
{
x = Find(x);
y = Find(y);
if(x != y)
{
par[x] = y;
}
}
void add(int u, int v)
{
edge[edgecnt].to = v;
edge[edgecnt].next = head[u];
head[u] = edgecnt++;
}
void addque(int u, int v, int i)
{
que[quecnt].to = v;
que[quecnt].num = i;
que[quecnt].next = qhead[u];
qhead[u] = quecnt++;
}
inline int read()
{
int x=0,flag=0;
char ch=getchar();
if(ch=='-') flag=1;
while(ch<'0'||ch>'9')ch=getchar();
while(ch>='0'&&ch<='9')x*=10,x+=ch-'0',ch=getchar();
if(flag) return -x;
return x;
}
void Tarjan(int u, int fa)
{
for(int i = head[u]; ~i; i = edge[i].next)
{
int v = edge[i].to;
if(v != fa)
{
Tarjan(v, u);
Merge(v, u);
vis[v] = 1;
}
}
vis[u] = 1;
for(int i = qhead[u]; ~i; i = que[i].next)
{
int v = que[i].to;
if(vis[v])
{
que[i].lca = Find(v);
que[i^1].lca = que[i].lca;
ans[que[i].num] = que[i].lca;
}
}
}
int main()
{
//ios_base::sync_with_stdio(0);
//cin.tie(0); cout.tie(0);
n=read(),q=read(),root=read();
//cin >> n >> q >> root;
memset(head, -1, sizeof(head));
memset(qhead, -1, sizeof(qhead));
for(int i = 0; i <= n; i++) par[i] = i;
int u, v;
for(int i = 0; i < n - 1; i++)
{
u = read(), v = read();
//cin >> u >> v;
add(u, v);
add(v, u);
}
for(int i = 0; i < q; i++)
{
u = read(), v = read();
//cin >> u >> v;
addque(u, v, i);
addque(v, u, i);
}
Tarjan(root, -1);
for(int i = 0; i < q; i++)
{
printf("%d\n", ans[i]);
//cout << ans[i] << endl;
}
return 0;
}
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