使用Tarjan算法,求解树上任意两节点的最近公共祖先
#include <bits/stdc++.h>
#define io cin.tie(0), cout.tie(0), ios::sync_with_stdio(false)
#define LL long long
#define ULL unsigned long long
#define EPS 1e-8
#define INF 0x7fffffff
#define SUB -INF - 1
using namespace std;
const int N = 500005;
int fa[N], head[N], cnt, head_query[N], cnt_query, ans[N];
bool vis[N];
struct edge
{
int to, next, num;
} edge[2 * N], query[2 * N];
void init()
{
for (int i = 0; i < 2 * N; i++)
{
edge[i].next = -1;
head[i] = -1;
query[i].next = -1;
head_query[i] = -1;
}
cnt = 0;
cnt_query = 0;
}
void addedge(int u, int v)
{
edge[cnt].to = v;
edge[cnt].next = head[u];
head[u] = cnt++;
}
void add_query(int x, int y, int num)
{
query[cnt_query].to = y;
query[cnt_query].num = num;
query[cnt_query].next = head_query[x];
head_query[x] = cnt_query++;
}
int find_set(int x) { return fa[x] == x ? x : find_set(fa[x]); }
void tarjan(int x)
{
vis[x] = true;
for (int i = head[x]; ~i; i = edge[i].next)
{
int y = edge[i].to;
if (!vis[y])
{
tarjan(y);
fa[y] = x;
}
}
for (int i = head_query[x]; ~i; i = query[i].next)
{
int y = query[i].to;
if (vis[y])
ans[query[i].num] = find_set(y);
}
}
int main()
{
init();
memset(vis, 0, sizeof(vis));
int n, m, root;
scanf("%d%d%d", &n, &m, &root);
for (int i = 1; i < n; i++)
{
fa[i] = i;
int u, v;
scanf("%d%d", &u, &v);
addedge(u, v);
addedge(v, u);
}
fa[n] = n;
for (int i = 1; i <= m; i++)
{
int a, b;
scanf("%d%d", &a, &b);
add_query(a, b, i);
add_query(b, a, i);
}
tarjan(root);
for (int i = 1; i <= m; i++)
printf("%d\n", ans[i]);
return 0;
}